Chapter 1.2, Problem 16E

Solution

To determine: the domain of the function algebraically.

The domain of the function is $(-\infty ,-4]\cup \left\{0\right\}\cup [4,\infty )$ .

Given Information:

Consider the function $\sqrt{x^4-16x^2}$ .

Concept used:

The expression under a radical cannot be negative.

Calculation:

For the domain of the function, the numerator has to be non-negative.

Simplify the function $\sqrt{x^4-16x^2}$ as follows.

  $\begin{array}{c}{x}^4-16x^2\ge 0\\ x^2\left(x^2-16\right)\ge 0\\ x^2\left(x+4\right)\left(x-4\right)\ge 0\\ x^2\ge 0,\left(x-4\right)\ge 0,\left(x+4\right)\ge 0\end{array}$

Let $x^2\ge 0$ .

It is clear that the domain is {0}.

Let $x-4\ge 0$ .

  $\begin{array}{c}x-4\ge 0\\ x\ge 4\end{array}$

The domain is $[4,\infty )$ .

Let $x+4\ge 0$ .

  $\begin{array}{c}x+4\ge 0\\ x\ge -4\end{array}$

The domain is $(-\infty ,-4]$ .

Therefore, the domain of the function is $(-\infty ,-4]\cup \left\{0\right\}\cup [4,\infty )$ .