# How do you find the local max and min for f(x) = x^3 - 27x?

Jan 9, 2016

Maximum at $x = - 3$, minimum at $x = 3$

#### Explanation:

Find the critical values of $f \left(x\right)$.

A critical value $c$ occurs when $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

Find $f ' \left(x\right)$ and set it equal to $0$.

$f \left(x\right) = {x}^{3} - 27 x$
$f ' \left(x\right) = 3 {x}^{2} - 27$

$3 {c}^{2} - 27 = 0$
$3 {c}^{2} = 27$
${c}^{2} = 9$
$c = \pm 3$

We know that two critical values, at which maxima or minima could occur, are $- 3$ and $3$. Since there are no values for which $f ' \left(x\right)$ is undefined, $- 3$ and $3$ are the only critical values.

We can use either or the first or second derivative test to determine if these are minima or maxima.

First Derivative Test

Examine the change in the function surrounding the critical values.

$f ' \left(- 4\right) = 21 \leftarrow \text{increasing}$
$f ' \left(- 3\right) = 0$
$f ' \left(- 2\right) = - 15 \leftarrow \text{decreasing}$

Since the derivative changes from increasing to decreasing when $x = - 3$, there is a relative maximum at $x = - 3$.

$f ' \left(2\right) = - 15 \leftarrow \text{decreasing}$
$f ' \left(3\right) = 0$
$f ' \left(4\right) = 21 \leftarrow \text{increasing}$

Since the derivative changes from decreasing to increasing when $x = 3$, there is a relative minimum at $x = 3$.

Second Derivative Test

Examine the concavity at each point to determine whether a minimum or maximum should occur.

First, find $f ' ' \left(x\right)$.

$f ' \left(x\right) = 3 {x}^{2} - 27$
$f ' ' \left(x\right) = 6 x$

Now, find the concavity at each of the critical values.

$f ' ' \left(- 3\right) = - 18$

Since $- 18 < 0$, the function is concave down. This means there will be a relative maximum when $x = 3$.

$f ' ' \left(3\right) = 18$

Since $18 > 0$, the function will be concave up. This means there will be a relative minimum when $x = 3$.