# What are the extrema of f(x)=x^3-2x+5  on [-2,2]?

Feb 4, 2016

Minimum: $f \left(- 2\right) = 1$
Maximum: $f \left(+ 2\right) = 9$

#### Explanation:

Steps:

1. Evaluate the endpoints of the given Domain

$f \left(- 2\right) = {\left(- 2\right)}^{3} - 2 \left(- 2\right) + 5 = - 8 + 4 + 5 = \textcolor{red}{1}$
$f \left(+ 2\right) = {2}^{3} - 2 \left(2\right) + 5 = 8 - 4 + 5 = \textcolor{red}{9}$

2. Evaluate the function at any critical points within the Domain.

To do this find the point(s) within the Domain where $f ' \left(x\right) = 0$
$f ' \left(x\right) = 3 {x}^{2} - 2 = 0$
$\rightarrow {x}^{2} = \frac{2}{3}$
$\rightarrow x = \sqrt{\frac{2}{3}} \text{ or } x = - \sqrt{\frac{2}{3}}$
$f \left(\sqrt{\frac{2}{3}}\right) \approx \textcolor{red}{3.9}$ (and, no, I didn't figure this out by hand)
f(-sqrt(2/3))~color(red)(~6.1)#

Minimum of $\left\{\textcolor{red}{1 , 9 , 3.9 , 6.1}\right\} = 1$ at $x = - 2$
Maximum of $\left\{\textcolor{red}{1 , 9 , 3.9 , 6.1}\right\} = 9$ at $x = + 2$

Here is the graph for verification purposes:
graph{x^3-2x+5 [-6.084, 6.4, 1.095, 7.335]}