Problem 52 - Intermediate Algebra

Question

Each of the following functions is one-to-one. Find the inverse of each function and express it using \(f^{-1}(x)\) notation. $$ f(x)=\frac{3}{x^{3}}-1 $$

Step-by-Step Solution

Answer

The inverse of the function is \( f^{-1}(x) = \left( \frac{3}{x + 1} \right)^{1/3} \).

Step 1: Understand the Function

We have the function \( f(x) = \frac{3}{x^{3}} - 1 \). Our goal is to find its inverse, which is a function such that when it is composed with \( f(x) \), it yields the identity function \( x \).

Step 2: Replace f(x) with y

Set \( y = f(x) = \frac{3}{x^{3}} - 1 \). We do this substitution to ease the algebraic manipulation needed to find the inverse.

Step 3: Solve for x in terms of y

To find the inverse function, we need to express \( x \) in terms of \( y \). Start by isolating the term with \( x \):\[ y + 1 = \frac{3}{x^{3}} \]

Step 4: Rearrange to Express x

Rearrange the equation to solve for \( x^{3} \):\[ x^{3} = \frac{3}{y + 1} \]

Step 5: Solve for x

Take the cube root of both sides to solve for \( x \):\[ x = \left( \frac{3}{y + 1} \right)^{1/3} \]

Step 6: Express the Inverse Function

Replace \( x \) with \( f^{-1}(x) \) and \( y \) with \( x \) to express the inverse function:\[ f^{-1}(x) = \left( \frac{3}{x + 1} \right)^{1/3} \]

Key Concepts

One-to-one functions

A function is considered one-to-one when it maps each element of its domain to a unique element in its range. This means that no two different inputs will produce the same output. In mathematical terms, if \( f(a) = f(b) \), then \( a = b \) must hold true. One-to-one functions play a key role in finding inverses because only these types of functions have inverses that are also functions.

To verify if a function is one-to-one, we can employ the Horizontal Line Test. This test involves drawing horizontal lines through the graph of a function. If any horizontal line crosses the graph more than once, the function is not one-to-one. However, if each line intersects the graph at most once, then the function is one-to-one.

In working with inverse functions, understanding whether a function is one-to-one is crucial because the existence of an inverse function depends on this property.

Algebraic manipulation

Algebraic manipulation involves changing the form of an expression to simplify or solve equations. It is an essential skill for finding the inverse of a function. When we want to find the inverse, we typically need to express one variable in terms of another through a series of algebraic manipulations.

For the function \( f(x) = \frac{3}{x^3} - 1 \), we use a series of algebraic steps to isolate \( x \):

• Add 1 to both sides to start isolating the \( x \) term: \( y + 1 = \frac{3}{x^3} \).
• Next, multiply both sides by \( x^3 \) to eliminate the fraction: \( x^3(y + 1) = 3 \).
• Divide by \( y + 1 \) to solve for \( x^3 \): \( x^3 = \frac{3}{y + 1} \).
• Finally, apply the cube root to both sides to solve for \( x \): \( x = \left( \frac{3}{y + 1} \right)^{1/3} \).

These manipulations allow us to switch the roles of \( x \) and \( y \) and express the inverse function cleanly.

Identity function

The concept of the identity function is foundational in understanding inverses. The identity function \( I(x) \) is simply \( x \). When a function \( f \) and its inverse \( f^{-1} \) are composed, they should return the identity function. This means

• \( f(f^{-1}(x)) = I(x) = x \)
• \( f^{-1}(f(x)) = I(x) = x \)

This characteristic is crucial because it confirms the functions effectively "undo" each other.

In the exercise we solved, our goal was to find \( f^{-1}(x) \) such that when it's composed with \( f(x) \), it results in the identity function \( x \). The steps of establishing \( y = f(x) \), isolating \( x \), and switching (\( x, y \)) roles lead us toward this goal, ensuring that \( f \circ f^{-1} \) yields the exact input that went in, verifying our inverse is correct.