How do you determine whether the function #f^('')(x) = 36x^2-12# is concave up or concave down and its intervals?

Answer 1

You can use the second derivative test.

You can determine the intervals on which your function is concave up and concave down by using the second derivative test.

ALl you have to do is examine the behavior of the second derivative around inflexion points, which are points for which the second derivative is equal to zero.

In your case, you have

#f^('') = 36x^2 - 12#
Make #f^('')=0# and see how many inflexion points you have
#36x^2 - 12 = 0#
#12(3x^2 -1) = 0#
#x^2 = 1/3#

Take the square root of both sides to get

#sqrt(x^2) = sqrt(1/3) => x = +- sqrt(3)/3#
You have two inflexion points, #x = -sqrt(3)/3# and #x = sqrt(3)/3#, so you're going to look at three intervals
SInce you're dealing with a square number, the sign of #f^('')# will always be positive for values that belong to this interval.
This means that #f(x)# is concave up on this interval.
This time, the expression #(3x^2-1)# will be negative for values that belong to this interval. This means that #f(x)# will be concave down on this interval.
Once again, the square of #x# will ensure that the expression #(3x^2-1)# will always be positive on this interval.
So, your function will be concave up on #(-oo,-sqrt(3)/3) uu (sqrt(3)/3, +oo)# and concave down on #(-sqrt(3)/3, sqrt(3)/3)#.
The graph of #f(x)# will have two inflexion points at #(-sqrt(3)/3, f(-sqrt(3)/3))# and #(sqrt(3)/3, f(sqrt(3)/3))#.
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Answer 2

To determine whether the function ( f''(x) = 36x^2 - 12 ) is concave up or concave down and its intervals, we need to analyze the sign of the second derivative.

If ( f''(x) > 0 ) for a given interval, the function is concave up on that interval. If ( f''(x) < 0 ) for a given interval, the function is concave down on that interval.

To find the intervals where the function is concave up or down, we first need to find the critical points of ( f''(x) ) by setting ( f''(x) = 0 ) and solving for ( x ).

( f''(x) = 36x^2 - 12 )

Setting ( f''(x) = 0 ):

( 36x^2 - 12 = 0 )

Solving for ( x ):

( 36x^2 = 12 )

( x^2 = \frac{12}{36} = \frac{1}{3} )

( x = \pm \sqrt{\frac{1}{3}} )

The critical points are ( x = -\sqrt{\frac{1}{3}} ) and ( x = \sqrt{\frac{1}{3}} ).

Next, we choose test points in the intervals defined by these critical points and evaluate ( f''(x) ) at these test points to determine the sign of ( f''(x) ) in each interval.

For example, we can choose test points ( x = -1 ) and ( x = 1 ) to evaluate ( f''(x) ) in the intervals ( (-\infty, -\sqrt{\frac{1}{3}}) ) and ( (\sqrt{\frac{1}{3}}, \infty) ), respectively.

( f''(-1) = 36(-1)^2 - 12 = 36 - 12 = 24 > 0 )

( f''(1) = 36(1)^2 - 12 = 36 - 12 = 24 > 0 )

Therefore, ( f''(x) > 0 ) for all ( x ) in both intervals.

Thus, the function ( f''(x) = 36x^2 - 12 ) is concave up for all real numbers ( x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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