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

Feb 16, 2016

There is a local maximum of $41$ at $x = - 2$. For an explanation of how to arrive at such an answer through the first and second derivative tests, refer to the explanation below.

#### 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 $0$ or is undefined.

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

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 36 x - 3$
$f ' \left(x\right) = 6 {x}^{2} - 6 x - 36$

Now, we should find the critical values by setting $f ' \left(x\right) = 0$.

$6 {x}^{2} - 6 x - 36 = 0$

Divide the terms by $6$.

${x}^{2} - x - 6 = 0$
$\left(x - 3\right) \left(x + 2\right) = 0$
$x = 3 \text{ }$ or $\text{ } x = - 2$

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 $x = - 2$ and $x = 3$.

• If $f ' \left(x\right)$ switches from increasing to decreasing around a critical value, there is a local maximum at the critical value.
• If $f ' \left(x\right)$ switches from decreasing to increasing around a critical value, there is a local minimum at the critical value.

$\star$ Testing the sign of the derivative around $x = \textcolor{red}{- 2} :$

$f ' \left(\textcolor{red}{- 3}\right) = 6 {\left(- 3\right)}^{2} - 6 \left(- 3\right) - 36 = 36 \leftarrow \textcolor{b l u e}{\text{INCREASING}}$
$f ' \left(\textcolor{red}{- 2}\right) = 6 {\left(- 2\right)}^{2} - 6 \left(- 2\right) - 36 = 0$
$f ' \left(\textcolor{red}{- 1}\right) = 6 {\left(- 1\right)}^{2} - 6 \left(- 1\right) - 36 = - 24 \leftarrow \textcolor{b l u e}{\text{DECREASING}}$

Since the derivative changes from increasing to decreasing at $x = - 2$, there is a local maximum at $x = - 2$. This is the location of the local maximum. The actual local maximum is color(green)(f(-2)=color(black)(ulcolor(green)(41.

Since we can easily envision the graph of a cubic, we know that there should be a local minimum at $x = 3$. However, to be rigorous, we should still test the sign of the derivative around $x = 3$ to prove that it is not a local maximum.

$\star$ Testing the sign of the derivative around $x = \textcolor{red}{3} :$

$f ' \left(\textcolor{red}{2}\right) = 6 {\left(2\right)}^{2} - 6 \left(2\right) - 36 = - 24 \leftarrow \textcolor{b l u e}{\text{DECREASING}}$
$f ' \left(\textcolor{red}{3}\right) = 6 {\left(3\right)}^{2} - 6 \left(3\right) - 36 = 0$
$f ' \left(\textcolor{red}{4}\right) = 6 {\left(4\right)}^{2} - 6 \left(4\right) - 36 = 36 \leftarrow \textcolor{b l u e}{\text{INCREASING}}$

As we suspected it would, the derivative changes from decreasing to increasing, revealing a local minimum of $f \left(3\right) = - 84$ at $x = 3$.

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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.

$f ' \left(x\right) = 6 {x}^{2} - 6 x - 36$
$f ' ' \left(x\right) = 12 x - 6$

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.

$\star$ Testing the concavity at $x = \textcolor{red}{- 2} :$

$f ' ' \left(\textcolor{red}{- 2}\right) = 12 \left(- 2\right) - 6 = - 30 \leftarrow \textcolor{b l u e}{\text{CONCAVE DOWN}}$

Thus, there is a local maximum at $x = 2$.

$\star$ Testing the concavity at $x = \textcolor{red}{3} :$

$f ' ' \left(\textcolor{red}{3}\right) = 12 \left(3\right) - 6 = 30 \leftarrow \textcolor{b l u e}{\text{CONCAVE UP}}$

Thus, there is a local minimum at $x = 3$.

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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]}