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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![Title: Understanding Inverse Functions
---
## Example Problem: Finding the Inverse Function
Given the function \( f \), we aim to find its inverse. The function \( f(x) \) and its domain are provided as follows:
\[ f(x) = x^2 + 5x, \quad x \geq -\frac{5}{2} \]
We are tasked with finding an expression for the inverse function \( f^{-1}(x) \) and its domain:
\[ f^{-1}(x) = \quad\boxed{ \quad }\quad , \quad x \geq -\frac{25}{4} \]
---
### Explanation of Terms:
- **Inverse Function**: The inverse of a function \( f \) is a function that, when composed with \( f \), yields the identity function. In terms of \( x \), it means that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
- **Domain**: The set of all possible input values (x-values) for which the function is defined.
### Steps to Find the Inverse Function:
1. **Write the function \( y = f(x) \)**:
\[ y = x^2 + 5x \]
2. **Solve for \( x \) in terms of \( y \)**:
- Rewriting the equation:
\[ y = x^2 + 5x \]
- Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 5 \), and \( c = -y \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-y)}}{2(1)} \]
\[ x = \frac{-5 \pm \sqrt{25 + 4y}}{2} \]
3. **Select the appropriate solution based on the domain**:
- Since \( x \geq -\frac{5}{2} \), we take the positive root:
\[ x = \frac{-5 + \sqrt{25 + 4y}}{2} \]
4. **Rewrite the solved equation by replacing \( y \) with \( x \) for](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24704420-4cdb-4013-8f71-69b6208e1345%2F280bcc3c-5c87-4cae-abb5-f3c4f735a3f8%2Fo6679ci_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Understanding Inverse Functions
---
## Example Problem: Finding the Inverse Function
Given the function \( f \), we aim to find its inverse. The function \( f(x) \) and its domain are provided as follows:
\[ f(x) = x^2 + 5x, \quad x \geq -\frac{5}{2} \]
We are tasked with finding an expression for the inverse function \( f^{-1}(x) \) and its domain:
\[ f^{-1}(x) = \quad\boxed{ \quad }\quad , \quad x \geq -\frac{25}{4} \]
---
### Explanation of Terms:
- **Inverse Function**: The inverse of a function \( f \) is a function that, when composed with \( f \), yields the identity function. In terms of \( x \), it means that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
- **Domain**: The set of all possible input values (x-values) for which the function is defined.
### Steps to Find the Inverse Function:
1. **Write the function \( y = f(x) \)**:
\[ y = x^2 + 5x \]
2. **Solve for \( x \) in terms of \( y \)**:
- Rewriting the equation:
\[ y = x^2 + 5x \]
- Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 5 \), and \( c = -y \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-y)}}{2(1)} \]
\[ x = \frac{-5 \pm \sqrt{25 + 4y}}{2} \]
3. **Select the appropriate solution based on the domain**:
- Since \( x \geq -\frac{5}{2} \), we take the positive root:
\[ x = \frac{-5 + \sqrt{25 + 4y}}{2} \]
4. **Rewrite the solved equation by replacing \( y \) with \( x \) for
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