# What are the extrema of f(x) = 64-x^2 on the interval [-8,0]?

Dec 30, 2015

Find the critical values on the interval (when $f ' \left(c\right) = 0$ or does not exist).

$f \left(x\right) = 64 - {x}^{2}$

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

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

$- 2 x = 0$

$x = 0$

And $f ' \left(x\right)$ is always defined.

To find the extrema, plug in the endpoints and the critical values. Notice that $0$ fits both of these criteria.

$f \left(- 8\right) = 0 \leftarrow \text{absolute minimum}$

$f \left(0\right) = 64 \leftarrow \text{absolute maximum}$

graph{64-x^2 [-8, 0, -2, 66]}