Problem 70 - A Graphical Approach to College Algebra

Question

The given function \(f\) is one-to-one. Find \(f^{-1}(x)\). $$f(x)=\frac{4-x}{5 x}$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \frac{4}{5x + 1}\).

Step 1: Substitute y for f(x)

First, substitute \(y\) for \(f(x)\) in the given function so that it becomes more manageable. This gives you the equation \(y = \frac{4-x}{5x}\).

Step 2: Solve for x in Terms of y

To find the inverse function, solve the equation \(y = \frac{4-x}{5x}\) for \(x\) in terms of \(y\). Multiply both sides by \(5x\) to eliminate the fraction: \[5xy = 4 - x.\] Then, rearrange it to \[5xy + x = 4.\] Factor out \(x\) from the left side: \[x(5y + 1) = 4.\] Finally, solve for \(x\) by dividing both sides by \(5y + 1\): \[x = \frac{4}{5y + 1}.\]

Step 3: Substitute x with f^{-1}(x) and y with x

Now, substitute \(x\) with \(f^{-1}(x)\) and \(y\) with \(x\) to find the inverse function: \[f^{-1}(x) = \frac{4}{5x + 1}.\]

Step 4: Verify the Inverse Function

To ensure the inverse function is correct, show that \(f(f^{-1}(x)) = x\). Substitute \(f^{-1}(x)\) into \(f(x)\): \[f\left(f^{-1}(x)\right) = f\left(\frac{4}{5x + 1}\right).\] Plug into the original function: \[= \frac{4 - \left(\frac{4}{5x + 1}\right)}{5\left(\frac{4}{5x + 1}\right)}.\] Simplify to show it evaluates to \(x\).

Key Concepts

One-to-One Function

A fundamental concept in mathematics is the idea of a one-to-one function. A function is defined as one-to-one, or injective, if different inputs always produce different outputs. In simpler terms, no two distinct inputs in the domain of the function share the same output. This property is crucial when determining whether a function has an inverse.
Consider the following practical aspects of one-to-one functions:

• Each element of the domain is paired with a unique element of the co-domain.
• A horizontal line drawn across the graph of a function intersects the graph at most once.
• Without being one-to-one, a function cannot have a proper inverse function.

When given a problem that involves finding an inverse function, confirming the one-to-one nature of the function sets the stage for solving the problem successfully.

Solving Equations

The process of solving equations is a cornerstone in the pursuit of understanding inverse functions. When tasked with finding the inverse of a function, one essentially solves an equation to express the original input variable in terms of the output variable. Let's break down the main actions involved:

• Switch the roles of the dependent and independent variables. In our example, this means substituting \( y \) for \( f(x) \).
• Rearrange the equation to isolate the input variable on one side. This often involves algebraic manipulations like factoring, expanding, and simplifying.

To illustrate, consider solving \( y = \frac{4-x}{5x} \). By carefully manipulating this equation, we identify \( x \) in terms of \( y \) as follows: multiply through by \( 5x \), rearrange, and factor out \( x \), culminating in a neat expression, \( x = \frac{4}{5y + 1} \). This critical step bridges the solution toward identifying the inverse function.

Function Verification

After deriving a potential inverse function, it's essential to verify that it is indeed correct. Verification lies in confirming that applying the original function to the inverse returns the initial input—this is expressed mathematically as \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). Such checks validate that the functions are genuine inverses.

• Plugging the inverse function into the original should simplify to \( x \).
• Substituting the original function into the inverse should also simplify back to \( x \).

For the function \( f(x) = \frac{4-x}{5x} \), the earlier derived inverse \( f^{-1}(x) = \frac{4}{5x + 1} \) must be plugged back into \( f \), and vice versa, to ensure the simplifications hold true. Verification provides confidence that the inverse function is mathematically correct and fulfills the necessary criteria of inverse operations.