# How do use the first derivative test to determine the local extrema #f(x)=x^3-2x +pi #?

Answer 1

See the explanation.

Find the critical numbers for #f#.

#f(x)=x^3-2x +pi #

#f'(x)=3x^2-2 #

#f'# is never undefined and is #f'(x) = 0# at #x= +- sqrt6/3#

On #(-oo, -sqrt6/3)#, we get #f'(x)# is positive
on #(-sqrt6/3, sqrt6/3)#, we get #f'(x)# is negative.

So #f# changes from increasing to decreasing as we move to the left past #x = -sqrt6/3#.
Therefore, #f(-sqrt6/3)# is a local maximum.

Recall: on #(-sqrt6/3, sqrt6/3)#, we get #f'(x)# is negative.
now, on #(sqrt6/3, oo)#, we get #f'(x)# is positive.
So #f# changes from decreasing to increasing as we move to the left past #x = sqrt6/3#.
Therefore, #f(sqrt6/3)# is a local minimum.

Calculating (simplifying) #f(-sqrt6/3)# and #f(sqrt6/3)# is left to the student.

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Answer 2

To use theTo use the firstTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1.To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal toTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zeroTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) andTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solveTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve forTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve forTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) toTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) toTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to findTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to findTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find criticalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find criticalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical pointsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical pointsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3.To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. SolveTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \):To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pmTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pmTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4.To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. TheseTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These criticalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical pointsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points areTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3.To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. EvaluateTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate theTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the signTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign ofTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) andTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) inTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in theTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervalsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determinedTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined byTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by theTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the criticalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical pointsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- ForTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5.To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x 0 \), so the functionTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function isTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasingTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) suchTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such thatTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- ForTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), forTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), chooseTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for exampleTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose aTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example,To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a testTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test pointTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point betweenTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between theTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the criticalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical pointsTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points,To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, likeTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- ForTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \).To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x >To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), chooseTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose aTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 >To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), soTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- ForTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x >To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), forTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for exampleTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example,To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), chooseTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose aTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a testTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test pointTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greaterTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater thanTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. DetermineTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine theTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the signTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign ofTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \),To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in eachTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each intervalTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- ForTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \).To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the functionTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function isTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasingTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) hasTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has aTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a localTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimumTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum atTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( xTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrtTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\fracTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \):To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}}To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \( fTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \( f'(To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \).
To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \( f'(0To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \).
-To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \( f'(0)To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \).
- TheTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point \( x_2 \) such that \( -\sqrt{\frac{2}{3}} < x_2 < \sqrt{\frac{2}{3}} \), for example, \( x_2 = 0 \).
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point \( x_3 \) such that \( x_3 > \sqrt{\frac{2}{3}} \), for example, \( x_3 = 1 \).
4. Determine the sign of \( f'(x) \) in each interval:
- For \( x < -\sqrt{\frac{2}{3}} \): \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \): \( f'(0) =To use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points.
3. Solve \( 3x^2 - 2 = 0 \) for \( x \): \( x^2 = \frac{2}{3} \) ⇒ \( x = \pm \sqrt{\frac{2}{3}} \).
4. These critical points are \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).
5. Test the sign of the derivative around each critical point to determine the behavior of the function:
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point less than \( -\sqrt{\frac{2}{3}} \), like \( x = -1 \). \( f'(-1) = 3(-1)^2 - 2 = 1 > 0 \), so the function is increasing.
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \), choose a test point between the critical points, like \( x = 0 \). \( f'(0) = 3(0)^2 - 2 = -2 < 0 \), so the function is decreasing.
- For \( x > \sqrt{\frac{2}{3}} \), choose a test point greater than \( \sqrt{\frac{2}{3}} \), like \( x = 1 \). \( f'(1) = 3(1)^2 - 2 = 1 > 0 \), so the function is increasing.
6. Based on the behavior of the derivative:
- The function \( f(x) \) has a local minimum at \( x = -\sqrt{\frac{2}{3}} \).
- The functionTo use the first derivative test to determine the local extrema of \( f(x) = x^3 - 2x + \pi \), follow these steps:
1. Find the first derivative of the function: \( f'(x) = 3x^2 - 2 \).
2. Set \( f'(x) \) equal to zero and solve for \( x \) to find critical points.
\[ 3x^2 - 2 = 0 \]
\[ 3x^2 = 2 \]
\[ x^2 = \frac{2}{3} \]
\[ x = \pm \sqrt{\frac{2}{3}} \]
3. Evaluate the sign of \( f'(x) \) in the intervals determined by the critical points.
- For \( x < -\sqrt{\frac{2}{3}} \), choose a test point \( x_1 \) such that \( x_1 < -\sqrt{\frac{2}{3}} \), for example, \( x_1 = -1 \).
- For \( -\sqrt{\frac{2}{3}} < x < \sqrt{\f