# How do you find the local max and min for  f(x)=x^3-2x+5 on the interval (-2,2)?

Jan 3, 2018

See below.

#### Explanation:

First find the 1st derivative of ${x}^{3} - 2 x + 5$

$\frac{d}{\mathrm{dx}} \left({x}^{3} - 2 x + 5\right) = 3 {x}^{2} - 2$

Maximum and minimum points have a horizontal gradient .i.e. gradient of $0$. Solving the 1st derivative will allow us ti identify these points.

$f ' \left(x\right) = 0$

$\therefore$

$3 {x}^{2} - 2 = 0 \implies x = \frac{\sqrt{6}}{3} , x = - \frac{\sqrt{6}}{3}$

Plugging these values into the second derivative, will allow us to find whether these are maximum, minimum or points of inflection. Using the following, if:

$f ' ' \left(x\right) > 0 \textcolor{w h i t e}{8888}$ minimum value

$f ' ' \left(x\right) < 0 \textcolor{w h i t e}{8888}$ maximum value

$f ' ' \left(x\right) = 0 \textcolor{w h i t e}{8888}$ maximum/minimum or point of inflection.

The second derivative is the derivative of the first derivative.

$\therefore$

$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 2\right) = 6 x$

Plugging in our values of $x$:

$6 \left(\frac{\sqrt{6}}{3}\right) = 2 \sqrt{6} \textcolor{w h i t e}{8888888}$ ( minimum value )

$6 \left(- \frac{\sqrt{6}}{3}\right) = - 2 \sqrt{6} \textcolor{w h i t e}{88}$ ( maximum value )

Both $\textcolor{w h i t e}{8888} x = \frac{\sqrt{6}}{3} , x = - \frac{\sqrt{6}}{3}$ are in $\left(- 2 , 2\right)$

GRAPH:

graph{y=x^3-2x+5 [-5, 5, -20, 20]}