# How do you find the critical numbers of y = abs(x^2 -1)?

May 7, 2016

The critical numbers are $- 1 , 0 , 1$

#### Explanation:

A critical number for $f$ is a number $c$, in the domain of $f$ at which $f ' \left(c\right)$ does not exist or $f ' \left(c\right) = 0$.

Therefore, we begin by finding $f ' \left(x\right)$ for $f \left(x\right) = \left\mid {x}^{2} - 1 \right\mid$.

NOte that the domain of $f$ is $\left(- \infty , \infty\right)$.

$f \left(x\right) = \left\mid {x}^{2} - 1 \right\mid = \left\{\begin{matrix}{x}^{2} - 1 & \text{if" & x^2-1 >= 0 \\ -(x^2-1) & "if} & {x}^{2} - 1 < 0\end{matrix}\right.$.

Investigating the sign of ${x}^{2} - 1$ shows that it is positive if $x < 1$ or $x > 1$ and it is negative if $- 1 < x < 1$.

Therefore,

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} - 1 & \text{if" & x <= -1 \\ -x^2+1 & "if" & -1 < x < 1 \\ x^2-1 & "if} & x \ge 1\end{matrix}\right.$.

Differentiating each piece gets us

$f ' \left(x\right) = \left\{\begin{matrix}2 x & \text{if" & x < -1 \\ -2x & "if" & -1 < x < 1 \\ 2x & "if} & x > 1\end{matrix}\right.$.

At the 'joints' of $x = - 1$ and $x = 1$ the left and right derivatives do not agree, so the derivative does not exist.

$- 1$ and $1$ are critical numbers for $f$.

Furthemore, $f ' \left(x\right) = 0$ at $x = 0$, so $0$ is also a critical number for $f$.

The critical numbers are $- 1 , 0 , 1$