Problem 29 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. \(f(x)=\frac{1}{x-1}, \quad x \neq 1\) \(g(x)=\frac{1}{x}+1, \quad x \neq 0\)

Step-by-Step Solution

Answer

Functions \(f\) and \(g\) are inverses because both \(f(g(x)) = x\) and \(g(f(x)) = x\).

Step 1: Understand the Inverse Function Property

The inverse function property states that two functions \(f\) and \(g\) are inverses of each other if \(f(g(x)) = x\) and \(g(f(x)) = x\) for every \(x\) in their domain. We need to verify both compositions to confirm they are inverses.

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

Substitute \(g(x)\) into \(f(x)\): \[f(g(x)) = f\left(\frac{1}{x} + 1\right) = \frac{1}{\left(\frac{1}{x} + 1\right) - 1}\]Simplify the expression:\[= \frac{1}{\frac{1}{x} + 1 - 1} = \frac{1}{\frac{1}{x}} = x\]So, \(f(g(x)) = x\) for \(x eq 0\).

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

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

Step 4: Conclusion

Both compositions \(f(g(x)) = x\) and \(g(f(x)) = x\) hold true for their respective domains. Hence, the functions \(f\) and \(g\) are inverses of each other.

Key Concepts

Inverse Function Property

The Inverse Function Property is a fundamental concept in mathematics used to determine if two functions are inverses of each other. For two functions, let's call them \( f \) and \( g \), to be inverses, they must satisfy two conditions:

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

This means that when you substitute one function into the other, you should end up back where you started, with \( x \). For example, if you take \( g(x) \) and use it as the input for \( f(x) \), you should simplify back to \( x \).
It is crucial that this property holds true for all values in the domain of \( g \) and \( f \) respectively. For our specific functions, \( f(x) = \frac{1}{x-1} \) and \( g(x) = \frac{1}{x} + 1 \), we find that \( f(g(x)) = x \) and \( g(f(x)) = x \) are valid in their respective domains. This demonstrates that \( f \) and \( g \) are indeed inverse functions of each other.

Function Composition

Function Composition is a process where one function is applied to the result of another function. If you have two functions, say \( f(x) \) and \( g(x) \), the composition \( f(g(x)) \) means substituting \( g(x) \) into \( f(x) \).
When checking if two functions are inverses, you compute two compositions: \( f(g(x)) \) and \( g(f(x)) \). The goal is to simplify each composition to just \( x \).

• In our case, \( f(g(x)) = \frac{1}{\left(\frac{1}{x} + 1\right) - 1} \), which simplifies to \( x \).
• Similarly, \( g(f(x)) = \frac{1}{\frac{1}{x-1}} + 1 \), which also simplifies to \( x \).

If both compositions result in \( x \), this confirms the functions are inverses over their respective domains. Function composition is a powerful tool in verifying the inverses of functions.

Domain of a Function

The domain of a function is the set of all possible input values (\( x \)) for which the function is defined. Understanding the domain is vital, especially when dealing with inverse functions.
For our functions:

• \( f(x) = \frac{1}{x-1} \) is defined for all real numbers except \( x = 1 \) as division by zero is undefined, making \( x eq 1 \) its domain.
• \( g(x) = \frac{1}{x} + 1 \) is defined for all real numbers except \( x = 0 \) for the same reason, giving it a domain of \( x eq 0 \).

When checking if \( f \) and \( g \) are inverses, their compositions must be valid within the domains of the respective functions. This is why the simplifications \( f(g(x)) = x \) and \( g(f(x)) = x \) only hold true for \( x eq 0 \) and \( x eq 1 \), respectively.
Ensuring that calculations respect the domain helps maintain the integrity of the results and confirms the functions are inverses across their valid inputs.