# How do you find the local max and min for  2x^3-3x^2-36x-3?

Jan 31, 2016

Local maximum of $41$ at $x = - 2$ and local minimum of $- 84$ at $x = 3$.

#### Explanation:

Here's the process, numbers and all:

Find $f ' \left(x\right)$ using the power rule.

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

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

The local minima and maxima will occur when the derivative equals $0$:

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

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

$\left(x - 3\right) \left(x + 2\right) = 0$

$x = 3 \text{ ",""" } x = - 2$

These are the two points at which maxima/minima could occur. There are three ways we can identify which is which:

The First Derivative Test:

Evaluate the sign of the first derivative around each local extremum.

Around $x = - 2$:

$f ' \left(- 3\right) = 36 \leftarrow \text{increasing}$
$f ' \left(- 1\right) = - 24 \leftarrow \text{decreasing}$

Since the first derivative switches from increasing to decreasing, there is a local maximum at $x = - 2$.

Around $x = 3$:

$f ' \left(2\right) = - 24 \leftarrow \text{decreasing}$
$f ' \left(4\right) = 36 \leftarrow \text{increasing}$

Since the first derivative switches from decreasing to increasing, there is a local minimum at $x = 3$.

The Second Derivative Test:

Find the second derivative of the function. Then, find the sign of it at each point.

• If $f ' ' \left(a\right) < 0$ and $f ' \left(a\right) = 0$, then there is a local maximum at $x = a$.

• If $f ' ' \left(a\right) > 0$ and $f ' \left(a\right) = 0$, then there is a local minimum at $x = a$.

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

$f ' ' \left(x\right) = 12 x - 6$

Determine the sign at each extremum.

$f ' ' \left(- 2\right) = - 30$

Since this is $< 0$, there is a local maximum at $x = - 2$.

$f ' ' \left(3\right) = 30$

Since this is $> 0$, there is a local minimum at $x = 3$.

Check a graph of the original function:

This method shouldn't be used as proof on a test, say, but it's a fine way to make sure you're on the right track with your answer.

graph{2x^3-3x^2-36x-3 [-5, 7, -120, 150]}

Indeed, there is a local maximum at $x = - 2$ and a local minimum at $x = 3$. Note that these are the locations of the extrema. The actual extrema are the function values of the points, that is, $f \left(- 2\right)$ and $f \left(3\right)$.

Thus, the local maximum is $f \left(- 2\right) = 41$ and the local minimum is $f \left(3\right) = - 84$.