Problem 19 - Algebra for College Students

Question

Verify that the two given functions are inverses of each other. $$ f(x)=5 x-9 \text { and } g(x)=\frac{x+9}{5} $$

Step-by-Step Solution

Answer

Yes, they are inverses because \(f(g(x)) = x\) and \(g(f(x)) = x\).

Step 1: Understand the Concept of Inverse Functions

Two functions, \(f\) and \(g\), are inverses of each other if and only if the composition of \(f(g(x))\) and \(g(f(x))\) equals \(x\). This means that applying one function and then the other should return the original input.

Step 2: Compute \(f(g(x))\)

Substitute \(g(x)\) into \(f(x)\). To do this, replace \(x\) in \(f(x) = 5x - 9\) with \(g(x) = \frac{x+9}{5}\).\[f(g(x)) = f\left(\frac{x+9}{5}\right) = 5\left(\frac{x+9}{5}\right) - 9\]Simplify the expression:\[f(g(x)) = x + 9 - 9 = x\]Thus, \(f(g(x)) = x\).

Step 3: Compute \(g(f(x))\)

Substitute \(f(x)\) into \(g(x)\). To do this, replace \(x\) in \(g(x) = \frac{x+9}{5}\) with \(f(x) = 5x - 9\).\[g(f(x)) = g(5x - 9) = \frac{5x - 9 + 9}{5}\]Simplify the expression:\[g(f(x)) = \frac{5x}{5} = x\]Thus, \(g(f(x)) = x\).

Step 4: Conclusion

Since both \(f(g(x)) = x\) and \(g(f(x)) = x\), the functions \(f(x) = 5x - 9\) and \(g(x) = \frac{x+9}{5}\) are indeed inverses of each other.

Key Concepts

Function Composition

Function composition is a crucial concept when we talk about inverse functions. It's like a mathematical chain reaction where the output of one function feeds directly into the input of another. Imagine you have two functions, \( f \) and \( g \). The composition \( f(g(x)) \) means you first apply \( g \) to \( x \), and whatever output you get, you then apply the function \( f \) to it.

When trying to determine if two functions are inverses, you need to test the compositions: \( f(g(x)) \) and \( g(f(x)) \). For them to be inverses, both compositions must simply return \( x \). If either composition results in anything other than \( x \), they are not inverses.

This concept is not only central to the idea of inverse functions but also widely used across different areas of mathematics, where compositions create complex functions out of simpler ones. By mastering function composition, you'll have a powerful tool in your mathematical toolkit.

Algebraic Manipulation

Algebraic manipulation is like solving a puzzle using various algebraic rules and techniques. It helps simplify expressions and solve equations by rearranging terms or applying arithmetic operations. Let's see how we use algebraic manipulation to verify inverse functions.

Consider our function \( f(x) = 5x - 9 \) and \( g(x) = \frac{x+9}{5} \). When we substitute \( g(x) \) into \( f(x) \), or \( f(g(x)) \), we replace each \( x \) in \( f \) with \( \frac{x+9}{5} \), and simplify the expression step by step, ensuring each algebraic operation keeps the expression equivalent. This results in \( f(g(x)) = x \).

Similarly, substituting \( f(x) \) into \( g(x) \) involves replacing \( x \) in \( g \) with \( 5x - 9 \). Again, simplifying through calculated algebraic steps gives \( g(f(x)) = x \).

Mastering these manipulations is essential for verifying inverses and solving many algebraic equations. Practicing various problems and observing patterns will improve your skills over time.

Mathematical Proofs

Mathematical proofs are like storytelling in math, where you logically show that something is true. They are essential for verifying mathematical truths, like proving that certain functions are inverses.

In our exercise, proving that \( f \) and \( g \) are inverses involves demonstrating two key compositions as equal to \( x \). Through detailed algebraic steps, we compute \( f(g(x)) = x \) and \( g(f(x)) = x \). These results form our proof.

• Step by step, we justify how each substitution and simplification maintains the truth of our equality.
• By following logical and clear reasoning, we eliminate any doubts or mistakes.

Proofs often require a balance between creativity and rigorous logic, making sure each statement is supported by previously known facts or rules. They help solidify our understanding and add depth to the knowledge of inverse functions.