1/x for x <-1 Let f(x) = { x² Vx for -1 1. (a) Evaluate f(-2) and f(2). (b) What is the range of f?
1/x for x <-1 Let f(x) = { x² Vx for -1 1. (a) Evaluate f(-2) and f(2). (b) What is the range of f?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem 22
Let \( f(x) \) be defined as follows:
\[
f(x) =
\begin{cases}
\frac{1}{x} & \text{for } x < -1 \\
x^2 & \text{for } -1 \leq x \leq 1 \\
\sqrt{x} & \text{for } x > 1
\end{cases}
\]
**(a) Evaluate \( f(-2) \) and \( f(2) \).**
**(b) What is the range of \( f \)?**
---
**Solution Approach:**
1. **Evaluating \( f(-2) \) and \( f(2) \):**
- For \( x = -2 \), since \(-2 < -1\), we use the first piece of the function: \( f(x) = \frac{1}{x} \).
\[ f(-2) = \frac{1}{-2} = -\frac{1}{2} \]
- For \( x = 2 \), since \( 2 > 1 \), we use the third piece of the function: \( f(x) = \sqrt{x} \).
\[ f(2) = \sqrt{2} \]
2. **Determining the range of \( f \):**
- For \( x < -1 \), \( f(x) = \frac{1}{x} \). As \( x \) approaches \(-\infty\), \( \frac{1}{x} \) approaches 0 from the negative side, and as \( x \) approaches \(-1\), \( \frac{1}{x} \) approaches \(-1 \).
Hence, the range for this part is \((-\infty, -1)\).
- For \(-1 \leq x \leq 1\), \( f(x) = x^2 \). Since the minimum value of \( x^2 \) is 0 (when \( x = 0 \)) and its maximum is 1 (when \( x = \pm 1 \)), the range for this part is \([0, 1]\).
- For \( x > 1 \), \( f(x) = \sqrt{x} \). As \( x \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F180e35fb-9b09-4fc5-8a6a-511bee286828%2F3c90fa22-a2a0-4f1a-b4a9-3a3b1cb89ed5%2Fp7koebr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 22
Let \( f(x) \) be defined as follows:
\[
f(x) =
\begin{cases}
\frac{1}{x} & \text{for } x < -1 \\
x^2 & \text{for } -1 \leq x \leq 1 \\
\sqrt{x} & \text{for } x > 1
\end{cases}
\]
**(a) Evaluate \( f(-2) \) and \( f(2) \).**
**(b) What is the range of \( f \)?**
---
**Solution Approach:**
1. **Evaluating \( f(-2) \) and \( f(2) \):**
- For \( x = -2 \), since \(-2 < -1\), we use the first piece of the function: \( f(x) = \frac{1}{x} \).
\[ f(-2) = \frac{1}{-2} = -\frac{1}{2} \]
- For \( x = 2 \), since \( 2 > 1 \), we use the third piece of the function: \( f(x) = \sqrt{x} \).
\[ f(2) = \sqrt{2} \]
2. **Determining the range of \( f \):**
- For \( x < -1 \), \( f(x) = \frac{1}{x} \). As \( x \) approaches \(-\infty\), \( \frac{1}{x} \) approaches 0 from the negative side, and as \( x \) approaches \(-1\), \( \frac{1}{x} \) approaches \(-1 \).
Hence, the range for this part is \((-\infty, -1)\).
- For \(-1 \leq x \leq 1\), \( f(x) = x^2 \). Since the minimum value of \( x^2 \) is 0 (when \( x = 0 \)) and its maximum is 1 (when \( x = \pm 1 \)), the range for this part is \([0, 1]\).
- For \( x > 1 \), \( f(x) = \sqrt{x} \). As \( x \)
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