Problem 30 - Algebra for College Students

Question

Determine whether \(f\) and \(g\) are inverse functions. $$ f(x)=\frac{1}{x+1} \text { and } g(x)=\frac{1-x}{x} $$

Step-by-Step Solution

Answer

No, \( f \) and \( g \) are not inverse functions.

Step 1: Understand the Concept of Inverse Functions

Two functions \( f \) and \( g \) are inverses of each other if \( f(g(x)) = x \) and \( g(f(x)) = x \) for all \( x \) in the domains of \( g \) and \( f \) respectively. This means applying \( g \) after \( f \) (and vice versa) returns the original input \( x \).

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

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

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

Substitute \( f(x) = \frac{1}{x+1} \) into \( g(x) = \frac{1-x}{x} \):\[g(f(x)) = g\left(\frac{1}{x+1}\right) = \frac{1 - \frac{1}{x+1}}{\frac{1}{x+1}}\]Simplify:\[g(f(x)) = \frac{\frac{x+1-1}{x+1}}{\frac{1}{x+1}} = x+1 \rightarrow ext{this does not simplify to } x\]

Step 4: Conclusion Based on Results

Since we have \( f(g(x)) = x \) for all permissible \( x \), but \( g(f(x)) eq x \), the functions \( f \) and \( g \) are not inverse functions of each other. For two functions to be inverses, both compositions need to simplify to \( x \).

Key Concepts

Function Composition

Function composition is like a two-step process where you perform one function and then another. When dealing with two functions, let's say \( f(x) \) and \( g(x) \), composing them is about applying one after the other. The notation \( f(g(x)) \) means you take the input \( x \), first apply \( g \) to it, and whatever that result is, you then input into \( f \).
For example, if \( g(x) = x + 1 \) and \( f(x) = x^2 \), then \( f(g(x)) \) becomes \( f(x+1) = (x+1)^2 \). Similarly, \( g(f(x)) \) would mean doing it the other way around: \( g\left(x^2\right) = x^2 + 1 \).
Function composition is key when we test if two functions are inverses.By checking both \( f(g(x)) \) and \( g(f(x)) \), we ensure both configurations return us back to our original \( x \).

Inverse Function Verification

Verifying inverse functions is like playing detective, confirming they indeed reverse each other perfectly. Two functions \( f \) and \( g \) are inverses if both composed versions return \( x \), meaning:

• \( f(g(x)) = x \)
• \( g(f(x)) = x \)

To verify whether \( f \) and \( g \) are inverses, we plug \( g(x) = \frac{1-x}{x} \) into \( f(x) = \frac{1}{x+1} \), and simplify to confirm if it equals \( x \). However, importantly, both directions need to work. This means after we simplify \( g(f(x)) \), it should also reduce back to \( x \).
If either direction fails to return \( x \), then the functions are not true inverses. This comprehensive check is crucial because inverse functions must perfectly undo what the other does.

Algebraic Manipulation

Algebraic manipulation is all about reshaping expressions to simplify or transform them. When verifying inverse functions, you'll need to use algebra to simplify expressions and check if they resolve to \( x \). This involves knowing how to handle fractions, expand terms, and manage complex expressions.
In our step-by-step solution, when finding \( f(g(x)) \), transforming \( \frac{1}{\left(\frac{1-x}{x}\right) + 1} \) into \( x \) required careful algebraic manipulation. Here, simplifying involved handling the fraction in the denominator and ensuring terms are correctly combined.
In \( g(f(x)) \), proper manipulation means simplifying \( \frac{1 - \frac{1}{x+1}}{\frac{1}{x+1}} \) and recognizing it does not result in \( x \), unraveling these tangled forms demands practice. Tools like converting complex fractions and diligently simplifying can make algebra much easier. Through practicing these skills, you can adeptly navigate and validate functional relationships.