Problem 31 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Definition of Inverse Functions

Two functions, \( f \) and \( g \), are inverse functions if their compositions yield the identity function, i.e., \( f(g(x)) = x \) and \( g(f(x)) = x \) for all \( x \) in the domains of the respective compositions.

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

We substitute \( g(x) \) into \( f(x) \), to find \( f(g(x)) \). Since \( g(x) = \frac{1}{x} \) and \( f(x) = x \), we have:\[ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{x}. \] This should equal \( x \) if \( f \) and \( g \) are inverse functions.

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

Next, substitute \( f(x) \) into \( g(x) \), to find \( g(f(x)) \). Since \( f(x) = x \) and \( g(x) = \frac{1}{x} \), we have:\[ g(f(x)) = g(x) = \frac{1}{x}. \] This should equal \( x \) if \( f \) and \( g \) are inverse functions.

Step 4: Analyze the Results

From Step 2, we have \( f(g(x)) = \frac{1}{x} \) instead of \( x \). From Step 3, \( g(f(x)) = \frac{1}{x} \) rather than \( x \). Therefore, neither composition results in the identity function on the respective domains.

Step 5: Conclusion

Since neither \( f(g(x)) \) nor \( g(f(x)) \) equals \( x \), \( f(x) = x \) and \( g(x) = \frac{1}{x} \) are not inverse functions of each other.

Key Concepts

Function Composition

Function composition refers to combining two functions to form a new function. Essentially, you substitute one function into another. This is denoted by \( f(g(x)) \) or \( g(f(x)) \), meaning you take the output of function \( g \) and use it as the input for function \( f \), and vice versa. Function composition helps us determine how one function interacts with another.

• When attempting to find if two functions, say \( f \) and \( g \), are inverse to each other, you must check the compositions \( f(g(x)) \) and \( g(f(x)) \).
• Neither of these operations should change the input value \( x \) if the functions are true inverses.

In our exercise, you substitute \( g(x) = \frac{1}{x} \) into \( f(x) = x \), and vice versa. The goal is to see if the result of these compositions returns the input \( x \). The results showed that they do not, confirming that the functions \( f \) and \( g \) are not inverses.

Identity Function

The identity function is a critical concept when determining inverse functions. It is written as \( I(x) = x \). This function always returns its input as the output, essentially leaving \( x \) unchanged. If a function composed with its potential inverse returns the identity function, they are considered inverses.

• For inverses, function compositions like \( f(g(x)) = x \) and \( g(f(x)) = x \) must hold true. This means that after performing the composition, you end up with the identity function.

In our case, neither \( f(g(x)) \) nor \( g(f(x)) \) resulted in \( x \), which means the identity function was not achieved. Therefore, \( f(x) = x \) and \( g(x) = \frac{1}{x} \) are not inverse functions since the identity function criterion was not met.

Domain of a Function

Understanding the domain of a function is crucial in evaluating function compositions and checking if functions are inverses. The domain of a function is the set of input values for which the function is defined. In simple terms, it's the collection of all \( x \) values you can plug into the function without causing any issues like division by zero.

• When evaluating function compositions for inverses, you must consider the domains of \( f \) and \( g \) to ensure that the compositions are valid. This is because an invalid input (like zero in a denominator) would break the function.
• For \( g(x) = \frac{1}{x} \), you must exclude \( x = 0 \) from its domain because the function would be undefined at that point.

While trying to determine if the two functions are inverses in the given exercise, the domains were considered. However, since the compositions \( f(g(x)) \) and \( g(f(x)) \) did not result in the identity function within the allowed domain, the functions were confirmed not to be inverses.