Problem 23 - Algebra for College Students

Question

Verify that the two given functions are inverses of each other. $$ \begin{aligned} &f(x)=\frac{1}{x-1} \quad \text { for } x>1 \quad \text { and } \\ &g(x)=\frac{x+1}{x} \quad \text { for } x>0 \end{aligned} $$

Step-by-Step Solution

Answer

Yes, the functions are inverses of each other.

Step 1: Understand the Definition of Inverse Functions

Two functions, \(f\) and \(g\), are inverses if \(f(g(x)) = x\) and \(g(f(x)) = x\) for all \(x\) within the domain of the functions. We will verify this property for the given functions.

Step 2: Find f(g(x))

Substitute \(g(x)\) into \(f(x)\): \[f(g(x)) = f\left(\frac{x+1}{x}\right) = \frac{1}{\frac{x+1}{x} - 1}.\]Simplify the expression:\[= \frac{1}{\frac{x+1}{x} - \frac{x}{x}} = \frac{1}{\frac{x+1-x}{x}} = \frac{1}{\frac{1}{x}} = x.\]Therefore, \(f(g(x)) = x\).

Step 3: Find g(f(x))

Substitute \(f(x)\) into \(g(x)\):\[g(f(x)) = g\left(\frac{1}{x-1}\right) = \frac{\frac{1}{x-1} + 1}{\frac{1}{x-1}}.\]Simplify the expression:\[= \frac{\frac{1+(x-1)}{x-1}}{\frac{1}{x-1}} = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = x.\]Therefore, \(g(f(x)) = x\).

Step 4: Conclusion on Inverses

Since \(f(g(x)) = x\) and \(g(f(x)) = x\) both hold true, the functions \(f(x)\) and \(g(x)\) are indeed inverses of each other for their respective domains.

Key Concepts

Function Composition

Function composition is like putting one function inside another to see how they work together. Imagine you have two functions: \( f(x) \) and \( g(x) \). Function composition is creating a new function by plugging \( g(x) \) into \( f(x) \), which is noted as \( f(g(x)) \). It transforms the input of one function using the other, and it's vital when working with inverse functions.

Let's simplify the process with an example. If \( g(x) = \frac{x+1}{x} \) and \( f(x) = \frac{1}{x-1} \), then to find \( f(g(x)) \), replace \( x \) in \( f(x) \) with \( g(x) \). This results in \( f(g(x)) = \frac{1}{\frac{x+1}{x} - 1} \).

• First, subtract 1 from \( g(x) \): \( \frac{x+1}{x} - 1 \)
• Next, simplify: \( = \frac{x+1-x}{x} = \frac{1}{x} \)
• Then, apply \( f(x) \): \( = \frac{1}{\frac{1}{x}} = x \).

The result, \( f(g(x)) = x \), confirms how these functions are mapped in the composition process.

Domain and Range

Understanding the domain and range is crucial when dealing with functions and their inverses. The domain is all possible values that can be input into a function, while the range is all the possible outputs.

For \( f(x) = \frac{1}{x-1} \), the domain excludes \( x = 1 \) because it would make the denominator zero, which is undefined. Therefore, \( x > 1 \). The range of \( f(x) \) would be all real numbers except 0, because the function never reaches zero.

Similarly, for \( g(x) = \frac{x+1}{x} \), the domain is \( x > 0 \) to avoid division by zero. The range is all real numbers except 1. These constraints come from not being able to output zero in \( g(x) \).

When verifying inverse functions, ensuring that each function's domain and range align correctly is important for each composition, \( f(g(x)) \) and \( g(f(x)) \), must lie within the domains of the respective functions involved.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is like clearing the fog in a math problem to see the solution path more clearly. It's all about performing algebraic manipulations that make expressions easier to interpret or compare.

Consider the expression \( f(g(x)) = \frac{1}{\frac{x+1}{x} - 1} \). To simplify:

• Subtract \( \frac{x}{x} = 1\) from \( \frac{x+1}{x} \), which gives \( \frac{1}{x} \).
• The result is \( \frac{1}{\frac{1}{x}} \), and flipping a fraction results in just \( x \).

Simplifying is especially vital when confirming inverse functions, because the equation must reduce neatly to \( x \) to prove that one function is the inverse of another.

This process isn't merely about reaching an answer; it's also about understanding each step, so you can solve any similar problems you encounter. Keep practicing these techniques to gain confidence in handling complex expressions.