# How do you find the critical numbers for f(x)= x^2 - 4 to determine the maximum and minimum?

Oct 27, 2016

#### Answer:

There is one critical point at $\left(0 , - 4\right)$ and it is a minimum.

#### Explanation:

$f \left(x\right) = {x}^{2} - 4$
$f \left(2\right) = {2}^{2} - 4 = 0$
$\therefore f ' \left(x\right) = 2 x$

At a max/min $f ' \left(x\right) = 0 \implies 2 x = 0 \implies x = 0$

And, $\therefore f ' ' \left(x\right) = 2 \implies f ' ' \left(2\right) > 0 \implies$minimum
graph{x^2-4 [-10, 10, -5, 5]}