Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![## Evaluating a Piecewise Function
Suppose that the function \( h \) is defined, for all real numbers, as follows:
\[
h(x) =
\begin{cases}
\frac{1}{2}x - 1 & \text{if } x \leq -2 \\
(x + 1)^2 + 2 & \text{if } -2 < x \leq 1 \\
\frac{1}{4}x - 2 & \text{if } x > 1
\end{cases}
\]
Find \( h(-3) \), \( h(0) \), and \( h(1) \).
\[
h(-3) = \, \_\_
\]
\[
h(0) = \, \_\_
\]
\[
h(1) = \, \_\_
\]
### Explanation Steps:
1. **For \( h(-3) \):** Use the first condition \( \frac{1}{2}x - 1 \) because \(-3 \leq -2\).
2. **For \( h(0) \):** Use the second condition \((x + 1)^2 + 2\) because \(-2 < 0 \leq 1\).
3. **For \( h(1) \):** Use the second condition \((x + 1)^2 + 2\) because \(-2 < 1 \leq 1\).
Input your answers into the box, then click "Check" to verify.
### Diagram Explanation
There are no graphs or diagrams provided with this problem. The problem is purely textual and involves evaluating expressions in a piecewise function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2afdb793-7012-417a-ba1a-2c4d4df308a5%2F52732ccf-6eed-450e-bf4e-d63a0cbe9647%2Fj7til5w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Evaluating a Piecewise Function
Suppose that the function \( h \) is defined, for all real numbers, as follows:
\[
h(x) =
\begin{cases}
\frac{1}{2}x - 1 & \text{if } x \leq -2 \\
(x + 1)^2 + 2 & \text{if } -2 < x \leq 1 \\
\frac{1}{4}x - 2 & \text{if } x > 1
\end{cases}
\]
Find \( h(-3) \), \( h(0) \), and \( h(1) \).
\[
h(-3) = \, \_\_
\]
\[
h(0) = \, \_\_
\]
\[
h(1) = \, \_\_
\]
### Explanation Steps:
1. **For \( h(-3) \):** Use the first condition \( \frac{1}{2}x - 1 \) because \(-3 \leq -2\).
2. **For \( h(0) \):** Use the second condition \((x + 1)^2 + 2\) because \(-2 < 0 \leq 1\).
3. **For \( h(1) \):** Use the second condition \((x + 1)^2 + 2\) because \(-2 < 1 \leq 1\).
Input your answers into the box, then click "Check" to verify.
### Diagram Explanation
There are no graphs or diagrams provided with this problem. The problem is purely textual and involves evaluating expressions in a piecewise function.
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