Problem 29 - Algebra and Trigonometry

Question

Use the Inverse Function Property to show that \(f\) and \(g\) are inverses of each other. $$ f(x)=\frac{1}{x} ; \quad g(x)=\frac{1}{x} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Inverse Function Property

For two functions, \( f(x) \) and \( g(x) \), to be inverses of each other, they must satisfy the conditions \( f(g(x)) = x \) and \( g(f(x)) = x \) for all \( x \) within the domain of the functions. Therefore, our goal is to verify both \( f(g(x)) = x \) and \( g(f(x)) = x \).

Step 2: Compute f(g(x))

First, calculate \( f(g(x)) \). By definition, \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x} \). So, substitute \( g(x) \) into \( f \): \[ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\left(\frac{1}{x}\right)} = x. \] Thus, \( f(g(x)) = x \).

Step 3: Compute g(f(x))

Next, calculate \( g(f(x)) \) by substituting \( f(x) \) into \( g \):\[ g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\left(\frac{1}{x}\right)} = x. \]Thus, \( g(f(x)) = x \).

Step 4: Conclusion

Since both conditions \( f(g(x)) = x \) and \( g(f(x)) = x \) are satisfied, \( f(x) \) and \( g(x) \) are indeed inverses of each other.

Key Concepts

Function Composition

Function composition is like a mix and match of functions, where one function's output becomes the input for another. To check if two functions are inverses, we use this concept to understand how they interact. Essentially, you're putting one function into the other. For functions \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x} \), their composition means plugging \( g(x) \) into \( f(x) \), and vice versa. This is how we verify if they truly undo each other.
When you compose functions, you perform the operations of one function followed by the other. Hence, function composition can be represented as \( (f \circ g)(x) = f(g(x)) \) and \( (g \circ f)(x) = g(f(x)) \).
To solve the original exercise, we compute both \( f(g(x)) \) and \( g(f(x)) \) and verify that both result in the original input \( x \). This tells us the functions are inverses and exemplifies the foundational role of function composition in understanding inverse functions.

Algebraic Functions

Algebraic functions are expressions formed using the basic arithmetic operations, which include addition, subtraction, multiplication, division, and exponentiation. The functions \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x} \) are simple algebraic functions, involving division. What makes exploring inverse functions interesting is how these simple operations can evolve into deeper mathematical concepts.
Understanding the inverse relationship between \( f \) and \( g \) in the exercise relies on manipulating algebraic expressions. We rewrite these expressions during function composition to see if they simplify back to \( x \).
With algebraic functions, it's important to remember properties like the reciprocal function, where the function and its inverse are the same. By calculating the same expression, both operations undo each other, as shown in the original solution.

Domain of Functions

The domain of a function is the set of all possible input values (\( x \)) for which the function produces a valid output. For \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x} \), the domain includes all real numbers except where the denominator is zero, because division by zero is undefined.
The key here is understanding that while a function might have an inverse, the domain must exclude values that cause undefined behavior. In our exercise, \( x eq 0 \). This restriction ensures that both \( f \) and \( g \) are well-defined for inverse operations.
Recognizing the correct domain is not just a formal step. It ensures the function behaves as expected. It's crucial for accurately confirming whether functions are truly inverses of each other. Without considering the domain, function composition or inverse calculation might lead to incorrect conclusions or undefined results.