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
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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