Chapter 5.2, Problem 5QR

To determine

To find : Where $f\left(x\right)=\sqrt{8-2x^2}$ is Differentiable.

Answer to Problem 5QR

The function is differentiable for $-2<x<2$ or $\left(-2,2\right)$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=\sqrt{8-2x^2}$

Calculation :

Consider the given equation:

  $f\left(x\right)=\sqrt{8-2x^2}$

Since,

Hence, $f\left(x\right)=\sqrt{8-2x^2}$ is defined for,

  $8-2x^2\ge 0\text{ (Square root of negative number is not defined)}$

Solve the above inequation for $x$ to get the domain of $f\left(x\right)=\sqrt{8-2x^2}$ ,

  $\begin{array}{l}8-2x^2\ge 0\\ 2x^2\le 8\\ x^2\le 4\end{array}$

Take square root on both sides and write the inequality as compound inequality.

  $\begin{array}{l}-\sqrt{4}\le \sqrt{x^2}\le \sqrt{4}\\ -2\le x\le 2\end{array}$

The domain of the function is $-2\le x\le 2$ or $\left[-2,2\right]$ .

Since the domain of the function is $-2\le x\le 2$ . Therefore, $f\left(x\right)$ is differentiable for all $x$ in the interior of its domain that is $-2<x<2$

Hence,

The function is differentiable for $-2<x<2$ or $\left(-2,2\right)$ .