Chapter 4.1, Problem 24E

To determine

To find : extreme values of the function and where they occur.

Answer to Problem 24E

The extreme value of the function is $-1$ and minimum at $x=0$ .

Explanation of Solution

Given information :

  $y=\frac{1}{x^2-1}$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The domain of the function is $\left(-\infty ,-1\right)\cup \left(-1,1\right)\cup \left(1,\infty \right)$ .

The function is not defined at $x=-1$ and, $x=1$ .

Differentiate $y=\frac{1}{x^2-1}$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{x^2-1}\right)\\ =\frac{-1}{{\left(x^2-1\right)}^2}\cdot \frac{d}{dx}\left(x^2-1\right)\\ =\frac{-1}{{\left(x^2-1\right)}^2}\cdot \left(2x\right)\\ =-\frac{2x}{{\left(x^2-1\right)}^2}\end{array}$

Take $\frac{dy}{dx}=0$ for critical points,

  $\begin{array}{c}-\frac{2x}{{\left(x^2-1\right)}^2}=0\\ x=0\left[\begin{array}{l}{x}^2-1\ne 0\\ x\ne \pm 1\end{array}\right]\end{array}$

The critical point is $x=0$ .

Now plug point $x=0$ in $y=\frac{1}{x^2-1}$ .

  $\begin{array}{c}y=\frac{1}{0^2-1}\\ =\frac{1}{-1}\\ =-1\end{array}$

Examine whether the function is either maxima or minima. Check the surrounding values of critical point.

As $x$ moves away from $0$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=0$ .