Chapter 4.1, Problem 23E

To determine

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

Answer to Problem 23E

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

Explanation of Solution

Given information :

  $y=\sqrt{x^2-1}$

Formula needed :

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

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

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

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

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

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

The critical points are $x=0$ and $x=\pm 1$ but $x=0$ does not lies in the domain.

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

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

Next plug point $x=-1$ in $y=\sqrt{x^2-1}$ .

  $\begin{array}{c}y=\sqrt{{\left(-1\right)}^2-1}\\ =\sqrt{1-1}\\ =0\end{array}$

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

As $x$ moves away from $1$ on right side, the value of $y$ increases spontaneous.

As $x$ moves away from $-1$ on left side, the value of $y$ decreases spontaneous.

Thus, minimum occur at $x=\pm 1$ .