Q. 62 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 47– 62, find the domain of each function. 

$h\left(z\right)=\frac{\sqrt{z+3}}{z-2}$

Step-by-Step Solution

Answer

The domain of the given function is $\left\{z\right|z\le -3,z\ne 2\}$

Step 1. Given Information

In the given problems we have to find the domain of given function.

$h\left(z\right)=\frac{\sqrt{z+3}}{z-2}$

Step 2. The function h tells us to take the square root of z + 3 and divide this result by z - 2 .

This requires that

$z+3\ge 0$

Subtract 3 on both side and change the inequality

$$\begin{gathered} z+3-3\le 0-3 \\ z\le -3 \end{gathered}$$

And $z-2\ne 0$

Add 2 on both side

$$\begin{gathered} z-2+2\ne 0+2 \\ z\ne 2 \end{gathered}$$

Step 3. Combining these two restrictions,

The domain of h is

$\left\{z\right|z\le -3,z\ne 2\}$