Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\frac{x^{3}+2}{x^{3}+1} $$

Short answer

The inverse function is \( f^{-1}(x) = \left(\frac{2-x}{x-1}\right)^{1/3} \). Both compositions confirm \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \).

Step-by-step solution

Understand the Problem

We need to find the inverse function of \( f(x) = \frac{x^3 + 2}{x^3 + 1} \) and verify that applying the original function to its inverse results in the identity function.

Set Up the Inverse Equation

Start by letting \( y = f(x) \), so \( y = \frac{x^3 + 2}{x^3 + 1} \). The goal is to solve for \( x \) in terms of \( y \).

Cross-Multiply to Eliminate Denominator

Multiply both sides by \( x^3 + 1 \) to get \( y(x^3 + 1) = x^3 + 2 \).

Rearrange the Equation

Rearrange the equation to isolate terms involving \( x \): \( yx^3 + y = x^3 + 2 \).

Bring Terms Involving \( x^3 \) Together

Move all \( x^3 \) terms to one side: \( yx^3 - x^3 = 2 - y \).

Factor Out \( x^3 \)

Factor \( x^3 \) from the left side: \( x^3(y - 1) = 2 - y \).

Solve for \( x^3 \)

Solve for \( x^3 \): \( x^3 = \frac{2 - y}{y - 1} \).

Find \( x \) in Terms of \( y \)

Take the cube root of both sides to solve for \( x \): \( x = \left(\frac{2 - y}{y - 1}\right)^{1/3} \).

Write the Inverse Function

The inverse function is \( f^{-1}(x) = \left(\frac{2 - x}{x - 1}\right)^{1/3} \).

Verify by Substituting \( f(x) \, ext{into} \, f^{-1}(x) \)

We need to show \( f^{-1}(f(x)) = x \). Substitute \( f(x) = \frac{x^3 + 2}{x^3 + 1} \) into \( f^{-1} \): \[ f^{-1} \left(\frac{x^3 + 2}{x^3 + 1}\right) = \left(\frac{2 - \frac{x^3+2}{x^3+1}}{\frac{x^3+2}{x^3+1} - 1}\right)^{1/3}. \]After simplifying, it gives \( x \), as expected.

Verify by Substituting \( f^{-1}(x) \, ext{into} \, f(x) \)

Now substitute \( f^{-1}(x) = \left(\frac{2 - x}{x - 1}\right)^{1/3} \) into \( f \): \[ f\left(\left(\frac{2 - x}{x - 1}\right)^{1/3}\right) = \frac{\left(\frac{2-x}{x-1}\right) + 2}{\left(\frac{2-x}{x-1}\right) + 1} \]This resolves to \( x \), confirming the identity holds.

Key concepts

Function Composition

When we talk about function composition, we are combining two functions to see how they interact with one another. Think of it as putting one function inside another, much like fitting a shirt inside a coat. In mathematical terms, if we have a function \( f(x) \) and another function \( g(x) \), the composition of these is written as \( f(g(x)) \). The output of \( g(x) \) becomes the input for \( f(x) \). This is particularly useful when dealing with inverse functions, as it helps verify if one function truly undoes the effect of the other. In our exercise, we check both \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \). These verify that the composed functions return us to our original input, ensuring that they indeed act as inverses of each other.

Algebraic Manipulation

Algebraic manipulation is the skillful art of rearranging and simplifying expressions to reach a desired form. This involves a series of techniques like adding, subtracting, multiplying, and factoring. In our exercise, the key steps involved manipulating the equation \( y = \frac{x^3 + 2}{x^3 + 1} \) in order to solve for \( x \) in terms of \( y \). This required cross-multiplying to eliminate the denominator, and then rearranging terms to separate those involving \( x \). These steps must be carefully executed to ensure that each side of the equation remains balanced and accurate. Such manipulations are foundational to solving equations and finding inverses, making them essential tools in algebra.

Cube Roots

Cube roots are used when we need to find the number that, when multiplied by itself three times, gives the original number. It is written with the root symbol and a small 3, like \( \sqrt[3]{x} \). In our step-by-step solution, we ultimately required finding \( x \) by taking the cube root of \( \frac{2 - y}{y - 1} \). The cube root operation essentially reverses the cube operation, enabling us to retrieve the original base value. This concept plays a critical role when determining inverse functions with cubic relationships, as it allows the isolation of variables in expressions where they have been raised to the third power.

Identity Function

An identity function is a simple function that returns exactly what you give it, symbolized mathematically as \( f(x) = x \). It's a baseline case in the world of functions. In our scenario, verifying that \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \) is evidence of the identity function at play. These verifications guarantee that despite applying the function and its inverse, the start and end values for \( x \) remain the same, embodying the essence of an identity function. This concept underscores the idea that true inverse functions should, in essence, "cancel out," leaving us with our original input.