Chapter 2: Q. 62 (page 69)

In Problems 47– 62, find the domain of each function.
$h (z) = \frac{\sqrt{z + 3}}{z - 2}$

Short Answer

Expert verified

The domain of the given function is ${z | z \leq - 3, z \neq 2}$

Step by step solution

Step 1. Given Information

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

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

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

$z + 3 \geq 0$

Subtract 3 on both side and change the inequality

$z + 3 - 3 \leq 0 - 3 z \leq - 3$

And $z - 2 \neq 0$

Add 2 on both side

$z - 2 + 2 \neq 0 + 2 z \neq 2$

Step 3. Combining these two restrictions,

The domain of h is

${z | z \leq - 3, z \neq 2}$

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In Problems 47– 62, find the domain of each function.
$h (z) = \frac{\sqrt{z + 3}}{z - 2}$

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. Combining these two restrictions,

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In Problems 47– 62, find the domain of each function.h(z)=z+3z-2

Short Answer

Step by step solution

Step 1. Given Information

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

Step 3. Combining these two restrictions,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Pure Maths

Calculus

Mechanics Maths

Probability and Statistics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In Problems 47– 62, find the domain of each function.
$h (z) = \frac{\sqrt{z + 3}}{z - 2}$