# How do you use the first and second derivatives to sketch f(x) = | (x^2) -1 |?

Apr 22, 2017

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

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

#### Explanation:

Graphing the function will help to answer the question:

So we can write the function as:

$f \left(x\right) = | {x}^{2} - 1 |$

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

$\text{ } = \left\{\begin{matrix}{x}^{2} - 1 & \null & x < - 1 \\ 1 - {x}^{2} & \null & - 1 < x < 1 \\ {x}^{2} - 1 & \null & x > 1\end{matrix}\right.$

Note that although $f \left(x\right)$ is continuous at $x = \pm 1$ the first and second derivatives are not defined at those points.

So then we can easily differentiate to get the first derivative:

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

And the second derivative is:

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