Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\left(\frac{x-1}{x+1}\right)^{3} $$

Short answer

The inverse function is \( f^{-1}(x) = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} \).

Step-by-step solution

Set Up Equation

To find the inverse function, denote the function as \( y = f(x) = \left( \frac{x-1}{x+1} \right)^3 \). Replace \( y \) with \( x \) and solve for \( x \) in terms of \( y \).

Remove Exponent

Take the cube root of both sides to eliminate the exponent: \( \sqrt[3]{x} = \frac{u-1}{u+1} \).

Solve for Inverse Function

Cross-multiply to eliminate the fraction: \( u - 1 = u (\sqrt[3]{x}) + \sqrt[3]{x} \). Rearrange to isolate \( u \): \( u(1 - \sqrt[3]{x}) = 1 + \sqrt[3]{x} \) and then solve for \( u \): \( u = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} \). Thus, the inverse function is \( f^{-1}(x) = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} \).

Verify \(f^{-1}(f(x))=x\)

Substitute \( f(x) = \left( \frac{x-1}{x+1} \right)^3 \) into our inverse: \( f^{-1}(f(x)) = f^{-1}((\frac{x-1}{x+1})^3) = \frac{1 + \sqrt[3]{(\frac{x-1}{x+1})^3}}{1 - \sqrt[3]{(\frac{x-1}{x+1})^3}} = x \), as expected.

Verify \(f(f^{-1}(x))=x\)

Substitute \( f^{-1}(x) = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} \) into \( f(x) \): \( f(f^{-1}(x)) = \left( \frac{\frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} - 1}{\frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}} + 1} \right)^3 = x \), thus confirming \( f(f^{-1}(x)) = x \).

Key concepts

Function Composition

When we talk about function composition, we are discussing the process of combining two functions to form a new one. This is much like plugging one function into another. If you have two functions, say \( f(x) \) and \( g(x) \), the composition is often written as \( f(g(x)) \). It means we first apply the function \( g \) to \( x \), and then take the result and plug it into the function \( f \).

In our scenario, we're looking at the compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \), aiming to show both are equal to \( x \). This composition helps verify that two functions are inverses of each other. In simpler terms, each function undoes the action of the other—like hitting the undo button on a computer. When you compose a function with its inverse, you should be back to where you started.

• \( f(f^{-1}(x)) = x \) implies that when \( f^{-1} \) modifies \( x \), \( f \) will bring it back to \( x \).
• \( f^{-1}(f(x)) = x \) shows us a similar relationship but starting with \( f(x) \) instead. Each function reverses the effects of the other.

Understanding function composition is essential because it forms the basis for checking if two functions are true inverses.

Inverse Verification

Inverse verification is the process of confirming that a given function truly is the inverse of another. For two functions \( f \) and \( f^{-1} \), if they are indeed inverses, substituting one into the other should give us back the original variable that was lost in transformation. In our problem, we have verified the inverse function through two checks: \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \). Let's see how this works:

• The first check, \( f^{-1}(f(x)) = x \), involves substituting the output of \( f(x) \) into \( f^{-1} \). Here, by calculating \( f^{-1}((\frac{x-1}{x+1})^3) \), we verified that the expression returns to \( x \).
• The second, \( f(f^{-1}(x)) = x \), means taking the result from \( f^{-1}(x) \) and applying \( f \). This effectively checks whether \( f(\frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}}) \) leads us back to just \( x \).

With both criteria satisfied, the inverse function has been thoroughly verified. This process confirms not only mathematical accuracy but reinforces our understanding of how inverse functions work in practice.

Cubic Functions

Cubic functions are polynomial functions of degree three and can often appear as \( f(x) = ax^3 + bx^2 + cx + d \). The unique aspect of cubic functions is their potential to have complex roots and inflection points, creating an intriguing shape for their graphs.

The exercise problem deals with a neat composition of functions where a cubic term is found after expressing \( f(x) = \left( \frac{x-1}{x+1} \right)^3 \). Here's why cubics are important in this context:

• With the cubing aspect, single changes in \( x \) can lead to more drastic shifts, unlike linear functions. This property makes working with them more nuanced, especially when finding inverses.
• In dealing with inverse functions, understanding cubes and cube roots becomes crucial to simplifying back to \( x \), as seen in the steps where cube roots needed to be calculated to solve for the inverse.

Comprehending cubic functions helps us tackle not only inverses but a variety of real-world mathematical modeling problems, as these functions can represent complex behavior beyond simple linearity.