Problem 21 - Contemporary Precalculus

Question

Use algebra to find the inverse of the given one-to-one function. $$f(x)=\frac{x^{3}-1}{x^{3}+5}$$

Step-by-Step Solution

Answer

Based on the step-by-step solution, use the following short answer: To find the inverse of a given one-to-one function, $$f(x) = \frac{x^3 - 1}{x^3 + 5}$$, we replace $$f(x)$$ with $$y$$, swap $$x$$ and $$y$$, and solve for $$y$$. After algebraic manipulation, the inverse function, $$f^{-1}(x)$$, is found to be $$f^{-1}(x) = \sqrt[3]{\frac{-1 - 5x}{x - 1}}$$.

Step 1: Understand the problem

The goal of this problem is to find the inverse function of $$f(x)$$ by algebraic means, meaning we want to find a function $$f^{-1}(x)$$ such that $$f(f^{-1}(x)) = x$$ for all x in the domain of $$f^{-1}$$ and $$f^{-1}(f(x)) = x$$ for all x in the domain of f. In other words, we want to find a function that "undoes" the function $$f(x)$$.

Step 2: Replace $$f(x)$$ with $$y$$

To make the algebraic manipulation easier, let's replace the function $$f(x)$$ with a variable y, so we now have: $$y = \frac{x^{3}-1}{x^{3}+5}$$

Step 3: Swap $$x$$ and $$y$$

To find the inverse function, we need to swap $$x$$ and $$y$$, then solve for $$y$$: $$x = \frac{y^{3}-1}{y^{3}+5}$$

Step 4: Solve for $$y$$

Now, let's solve for $$y$$ algebraically. First, we'll get rid of the fraction by multiplying both sides with $$y^3 + 5$$: $$(x)(y^{3}+5) = y^{3}-1$$ Now, distribute $$x$$ on the left side: $$xy^{3} + 5x = y^{3}-1$$ Move all terms involving $$y$$ to one side and all other terms to the other side: $$xy^{3} - y^{3} = -1 - 5x$$ Factor $$y^3$$ on the left side: $$y^{3}(x - 1) = -1 - 5x$$ Now, divide both sides by $$(x - 1)$$ to solve for $$y^3$$: $$y^{3} = \frac{-1 - 5x}{x - 1}$$ Finally, find the cube root of both sides to solve for $$y$$: $$y = \sqrt[3]{\frac{-1 - 5x}{x - 1}}$$

Step 5: Write the inverse function

Now that we have solved for $$y$$, we can write down the inverse function, $$f^{-1}(x)$$: $$f^{-1}(x) = \sqrt[3]{\frac{-1 - 5x}{x - 1}}$$ This is the inverse function, which undoes the action of the original function $$f(x)$$.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a critical skill in mathematics that allows us to rearrange and simplify expressions or equations. In this context, it's used to find the inverse of a function. To tackle this exercise with algebra, we start by renaming the function expression to a simpler variable, like using \(y\) instead of \(f(x)\).
This substitution simplifies the appearance of the equation and lays the groundwork for reversing the roles of \(x\) and \(y\). By swapping \(x\) and \(y\) in the equation, we're essentially setting up the problem to solve for the inverse function.

To isolate \(y\), we need to eliminate fractions, distribute terms, and move them around. Here's how:

• Multiply out any denominators to simplify the equation, making it easier to isolate the term you want to solve for.
• Combine like terms and factor common elements to further simplify the equation.
• Finally, perform operations such as adding, subtracting, multiplying, or dividing both sides of the equation as necessary to solve for the desired variable.

These steps of algebraic manipulation ensure clarity and precision, vital for finding exact solutions in inverse functions.

One-to-One Function

A one-to-one function is a function where each output is linked to exactly one unique input—no inputs produce the same output. This characteristic is crucial for functions to have inverses. If a function isn't one-to-one, it might not possess an inverse across its entire domain.
Mathematically, a function \(f(x)\) is one-to-one if, whenever \(f(a) = f(b)\), it follows that \(a = b\). This property simplifies into two main checks:

• Horizontal Line Test: On its graph, a one-to-one function will never be intersected by a horizontal line more than once.
• Monotonicity: The function is always increasing or decreasing, meaning it doesn't change direction more than once.

Because our original function \(f(x) = \frac{x^3 - 1}{x^3 + 5}\) has each distinct \(x\) mapping to a distinct \(f(x)\), it ensures the existence of an inverse function, confirming that mathematical reversals are legitimate and well-defined.

Cube Root

The cube root operation allows us to take a number \(n\) and find a value \(x\) such that \(x^3 = n\). It is the inverse operation of cubing a number and is indicated mathematically as \(\sqrt[3]{n}\). In the context of finding inverse functions, once we've found \(y^3\) as a part of our solution, the cube root gives us back \(y\).
To solve for \(y\), we take the cube root of the entire expression:

• Calculate: Apply \(\sqrt[3]{ }\) to both sides to isolate \(y\).
• Interpret: This operation allows complex expressions under the cube root—a manipulation useful when functions involve cubic transformations.

Thus, determining the cube root is essential not just for finding the inverse but also for simplifying and understanding the structure of cubic relationships within algebraic functions. As seen in the solution, applying the cube root is the last key step that consolidates our earlier algebraic manipulations into the clear form of the inverse function: \(f^{-1}(x) = \sqrt[3]{\frac{-1 - 5x}{x - 1}}\).