Problem 71 - College Algebra with Modeling and Visualization

Question

Find a formula for \(f^{-1}(x) .\) Identify the domain and range of \(f^{-1}\). Verify that \(f\) and \(f^{-1}\) are inverses. $$ f(x)=5 x-15 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{x + 15}{5} \), with domain and range as all real numbers.

Step 1: Understand the Problem

We are given a function \( f(x) = 5x - 15 \) and need to find the inverse function \( f^{-1}(x) \). We will also need to identify the domain and range of the inverse function and verify that \( f \) and \( f^{-1} \) are indeed inverses of each other.

Step 2: Find the Inverse Function

To determine \( f^{-1}(x) \), we first replace \( f(x) \) with \( y \):\[ y = 5x - 15 \]Next, solve for \( x \) as a function of \( y \):\[ y + 15 = 5x \]\[ x = \frac{y + 15}{5} \]Thus, the inverse function is \( f^{-1}(x) = \frac{x + 15}{5} \).

Step 3: Identify Domain and Range of Inverse Function

For the given linear function \( f(x) = 5x - 15 \), the domain is all real numbers. Consequently, the range is also all real numbers. Since the inverse function reverses the roles of domain and range, the domain and range of \( f^{-1}(x) \) are both all real numbers.

Step 4: Verify Functions are Inverses

To verify \( f \) and \( f^{-1} \) are inverses, show that:\[ f(f^{-1}(x)) = x \]and\[ f^{-1}(f(x)) = x \]1. Substitute \( f^{-1}(x) \) into \( f \): \[ f\left( \frac{x + 15}{5} \right) = 5 \left( \frac{x + 15}{5} \right) - 15 = x + 15 - 15 = x \]2. Substitute \( f(x) \) into \( f^{-1} \): \[ f^{-1}(5x - 15) = \frac{(5x - 15) + 15}{5} = \frac{5x}{5} = x \]Both compositions return \( x \), confirming they are inverses.

Key Concepts

Domain and Range in Inverse Functions

When dealing with inverse functions, understanding the domain and range is crucial. The domain of a function is the set of all possible input values, while the range is the set of all possible output values. In our exercise, we started with the function \( f(x) = 5x - 15 \). For a linear function like this, the domain is all real numbers, since you can plug any real number into \( x \).

As for the range, it's also all real numbers because every value can be achieved by plugging some \( x \) value into the function. However, when you find the inverse, these roles switch: the domain of the original function becomes the range of the inverse, and the range of the original function becomes the domain of the inverse.

Thus, the inverse function \( f^{-1}(x) = \frac{x + 15}{5} \) also has all real numbers for both its domain and its range. This reciprocal relationship is an essential property of inverse functions, helping to ensure they map effectively between input and output, just like their original counterparts.

Linear Functions and Their Inverses

Linear functions are among the simplest types of functions we encounter in mathematics. They are in the form \( f(x) = mx + b \), where \( m \) and \( b \) are constants. The function given in the exercise, \( f(x) = 5x - 15 \), is a classic linear function with a slope \( m = 5 \) and a y-intercept of \( -15 \).

To find the inverse of a linear function, you swap the roles of \( x \) and \( f(x) \), and then solve for \( x \) in terms of the new \( y \). This process results in the inverse function \( f^{-1}(x) \). For a linear function, this inverse will also be linear, maintaining a simple yet powerful transformation. In our case, the inverse function \( f^{-1}(x) = \frac{x + 15}{5} \) shows that for each step away from the origin (0,0), the function behaves predictably and proportionally, scaling inputs directly with clear relationships between \( x \) and \( y \).

Understanding linear functions and their inverses helps us solve real-world problems where relationships between variables are consistent and predictable.

Function Composition and Inverse Verification

Function composition is a key concept when working with inverse functions. It involves combining two functions to see how they interact. Specifically, we verify inverse functions by showing that the composition of a function and its inverse results in the identity function.

In simpler terms, if \( f \) and \( f^{-1} \) are true inverses, then inserting \( f^{-1} \) into \( f \) should return you to your original input, and vice versa. This is expressed mathematically as \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

For our exercise, substituting \( f^{-1}(x) = \frac{x + 15}{5} \) into \( f(x) = 5x - 15 \) results in simply \( x \), confirming \( f(f^{-1}(x)) = x \). Likewise, when we put \( f(x) \) into \( f^{-1}(x) \), we also end up with \( x \), affirming \( f^{-1}(f(x)) = x \).

These checks ensure the integrity of inverse functions and demonstrate how the process of composition helps us confirm the validity and the correctness of function inverses, assuring their place in mathematical equations and applications.