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

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Answer 1 · 2016-05-07T17:21:26

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

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

Therefore, we begin by finding $f'(x)$ for $f(x) = abs(x^2-1)$ .

NOte that the domain of $f$ is $(-oo,oo)$ .

$f(x) = abs(x^2-1) = {(x^2-1,"if",x^2-1 >= 0),(-(x^2-1),"if",x^2-1 < 0) :}$ .

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,

$f(x) = {(x^2-1,"if",x <= -1),(-x^2+1,"if",-1 < x < 1),(x^2-1,"if",x >= 1) :}$ .

Differentiating each piece gets us

$f'(x) = {(2x,"if",x < -1),(-2x,"if",-1 < x < 1),(2x,"if",x > 1) :}$ .

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

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

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