# What is the derivative of a function equal to at a critical point?

##### 1 Answer
Nov 4, 2015

It is either 0 or it does not exist.

#### Explanation:

A critical number for a function is a number in the domain of the function at which the derivative is either 0 or fails to exist.

Examples

$f \left(x\right) = {x}^{3} + 5 {x}^{2} - 7$ has derivative $f ' \left(x\right) = 3 {x}^{2} + 10 x$ and critical numbers $0$ and $- \frac{10}{3}$

$g \left(x\right) = \frac{x}{x - 5}$ has derivative $g ' \left(x\right) = \frac{- 5}{x - 5} ^ 2$ which is never $0$ and is undefined only at $5$ which is not in the domain of $g$. So, $g$ has no critical numbers.

$h \left(x\right) = {\left(x - 1\right)}^{\frac{2}{3}}$ has derivative $h ' \left(x\right) = \frac{2}{3 {\left(x - 1\right)}^{\frac{1}{3}}} = \frac{2}{3 \sqrt[3]{x - 1}}$ which is never $0$ and is undefined only at $1$ which is in the domain of $h$. So, $h$ has critical number $1$.