How do you find the critical points for abs(x^2 -1)?

Nov 26, 2015

They are $0 , 1 , \text{and } - 1$.

Explanation:

Let $f \left(x\right) = \left\mid {x}^{2} - 1 \right\mid$, so

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

At all $x$ other than $1$ and $- 1$, we can quickly see that $f ' \left(x\right) = 2 x$ or $- 2 x$

Now, ${\lim}_{x \rightarrow {1}^{+}} \frac{f \left(x\right) - f \left(1\right)}{x - 1} = {\lim}_{x \rightarrow {1}^{+}} \frac{{x}^{2} - 1}{x - 1} = 2$,

but ${\lim}_{x \rightarrow {1}^{-}} \frac{f \left(x\right) - f \left(1\right)}{x - 1} = {\lim}_{x \rightarrow {1}^{-}} \frac{- {x}^{2} + 1}{x - 1} = - 2$.

We conclude that $f$ has no derivative at $1$.

A similar argument will show that there is no derivative at $- 1$.

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

Because both $1$ and $- 1$ are in the domain of $f$ and $f '$ does not exists at either, these are both critical numbers for $f$.

In addition, $f ' \left(x\right) = 0$ at $x = 0$, so $0$ is another critical number for $f$.