Chapter 1.1, Problem 50E

To determine

To find: domain of $h\left(x\right)=\frac{x+2}{\sqrt{x^2-7}}$

Answer to Problem 50E

  $\left\{x|x>\sqrt{7}\right\}$

Explanation of Solution

Given information:

The function, $h\left(x\right)=\frac{x+2}{\sqrt{x^2-7}}$

Calculation:

For $h\left(x\right)$ , the denominator should not be zero as well as the radicand should not be negative. So, to evaluate the value of x , put the inequality $\sqrt{x^2-7}$ less than or equal to 0 as shown below.

  $\begin{array}{c}\sqrt{x^2-7}\le 0\\ x^2-7\le 0\left(\text{Squaring on both sides}\right)\\ x^2\le 7\left(\text{Adding 7 to both sides}\right)\\ x\le \sqrt{7}\left(\text{Taking square roots on both sides}\right)\end{array}$

The denominator of function $h\left(x\right)$ will be zero or negative at $x\le \sqrt{7}$ . So, this will not be included in the domain.

Hence, the domain will be $\left\{x|x>\sqrt{7}\right\}$ .