# Question #51aa2

Sep 21, 2016

#### Explanation:

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

Solving the inequality ${x}^{2} - 1 > 0$ gets us,

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

So the derivative is given by

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

$2 x = 1$ at $x = \frac{1}{2}$ but the derivative at $x = \frac{1}{2}$ is not given by $2 x$.

$- 2 x = 1$ at $x = - \frac{1}{2}$ and at $x = - \frac{1}{2}$, the derivative is given by $- 2 x$.

So the slope is $1$ and $x = - \frac{1}{2}$. At that point, we get $y = \frac{3}{4}$.

So the only point on the curve at which the slope is $1$ is $\left(- \frac{1}{2} , \frac{3}{4}\right)$.