# How do you find local maximum value of f using the first and second derivative tests: #2x^3-3x^2-36x-3#?

##### 1 Answer

There is a local maximum of

#### Explanation:

**1. Find the critical values of the function. These are where the local maximum could exist.**

A function's critical values exist where the function's derivative equals

To find this function's derivative, use the **power rule.**

#f(x)=2x^3-3x^2-36x-3#

#f'(x)=6x^2-6x-36#

Now, we should find the critical values by setting

#6x^2-6x-36=0#

Divide the terms by

#x^2-x-6=0#

#(x-3)(x+2)=0#

#x=3" "# or#" "x=-2#

These are the function's two critical values. These are where the local maxima or minima could exist.

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**2. Use the first derivative test to see if the function switches from increasing to decreasing or decreasing to increasing around the critical values.**

Test the sign of the derivative around the points at

- If
#f'(x)# switches from**increasing to decreasing**around a critical value, there is a**local maximum**at the critical value. - If
#f'(x)# switches from**decreasing to increasing**around a critical value, there is a**local minimum**at the critical value.

#f'(color(red)(-3))=6(-3)^2-6(-3)-36=36larrcolor(blue)("INCREASING")#

#f'(color(red)(-2))=6(-2)^2-6(-2)-36=0#

#f'(color(red)(-1))=6(-1)^2-6(-1)-36=-24larrcolor(blue)("DECREASING")#

Since the derivative changes from increasing to decreasing at *location* of the local maximum. The *actual* local maximum is

Since we can easily envision the graph of a cubic, we know that there should be a local minimum at

#f'(color(red)2)=6(2)^2-6(2)-36=-24larrcolor(blue)("DECREASING")#

#f'(color(red)3)=6(3)^2-6(3)-36=0#

#f'(color(red)4)=6(4)^2-6(4)-36=36larrcolor(blue)("INCREASING")#

As we suspected it would, the derivative changes from decreasing to increasing, revealing a local minimum of

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**3. Examine the concavity at each critical value through the second derivative (the second derivative test).**

First, find the second derivative by differentiating the first derivative.

#f'(x)=6x^2-6x-36#

#f''(x)=12x-6#

The process for the second derivative test is as follows:

- If the value of the second derivative at a critical value is
**positive,**the function is**concave up**at that point and there is a**local minimum**. - If the value of the second derivative at a critical value is
**negative,**the function is**concave down**at that point and there is a**local maximum**.

Note that we should get the same results that the first derivative test gave us--this is just another way of reaching the same conclusion.

#f''(color(red)(-2))=12(-2)-6=-30larrcolor(blue)("CONCAVE DOWN")#

Thus, there is a local maximum at

#f''(color(red)3)=12(3)-6=30larrcolor(blue)("CONCAVE UP")#

Thus, there is a local minimum at

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If you're able, always consult a graph of the function to check your answer:

graph{2x^3-3x^2-36x-3 [-5, 7, -106.8, 107]}