Problem 36 - 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. $$ f(x)=5 x-1 $$

Step-by-Step Solution

Answer

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

Step 1: Write the Given Function

We start with the given function \( f(x) = 5x - 1 \). The goal is to find the inverse function denoted by \( f^{-1}(x) \).

Step 2: Replace f(x) with y

Replace \( f(x) \) with \( y \) to simplify our calculations: \( y = 5x - 1 \).

Step 3: Solve for x in Terms of y

Rearrange the equation to solve for \( x \): Add 1 to both sides: \( y + 1 = 5x \). Next, divide both sides by 5 to isolate \( x \): \( x = \frac{y + 1}{5} \).

Step 4: Swap x and y to Find the Inverse

To find the inverse function, swap \( x \) and \( y \): \( y = \frac{x + 1}{5} \). This equation gives us \( f^{-1}(x) = \frac{x + 1}{5} \).

Step 5: Express the Inverse Function in Notation

Write the inverse function using \( f^{-1}(x) \) notation: \( f^{-1}(x) = \frac{x + 1}{5} \).

Key Concepts

One-to-One Functions

When trying to find the inverse of a function, it's important that the function is one-to-one. A one-to-one function is one where each output value has a unique input value. This ensures that no two different inputs map to the same output.
One special characteristic of one-to-one functions is that they pass the "horizontal line test." If a horizontal line intersects the graph of the function at more than one point, the function is not one-to-one.
This property is crucial because only one-to-one functions have inverse functions that are also functions. For example, our given function, \( f(x) = 5x - 1 \), is linear with a non-zero slope, which confirms it is one-to-one. This implies it has an inverse. Knowing this ahead of time simplifies finding the inverse and trusting it's a function.

Function Notation

Function notation helps us understand and communicate mathematical relationships clearly. When a function is expressed as \( f(x) \), \( f \) represents the function, and \( x \) is the variable input.
Given \( f(x) = 5x - 1 \), we understand that for any input \( x \), the function returns the value derived by multiplying \( x \) by 5 and then subtracting 1.
When working with inverse functions, such as \( f^{-1}(x) \), the notation indicates that we are looking at the process of reversing the function's operations. In the inverse, instead of applying the function to find a result from a given \( x \), we use the inverse to find the unknown \( x \) for a given result value.
Correct use of function notation helps keep track of inputs and outputs, making it pivotal when performing algebraic manipulations to find inverses.

Algebraic Manipulation

Algebraic manipulation is the technique of rearranging mathematical expressions and equations to find unknowns or simplify expressions. When finding an inverse function, like \( f(x) = 5x - 1 \), algebraic manipulation allows us to reverse the operations performed by the original function.
**Process of Finding Inverse Using Algebraic Manipulation:**

• Start by rewriting the function with \( y \) replacing \( f(x) \) to make operations clearer.
• Rearrange the equation \( y = 5x - 1 \) to isolate \( x \). We do this by first adding 1 to both sides, then dividing each side by 5.
• Once \( x \) is isolated, we switch the roles of \( y \) and \( x \) to express the inverse. This results in \( y = \frac{x + 1}{5} \).
• The final step is to rewrite the equation using inverse function notation, \( f^{-1}(x) = \frac{x + 1}{5} \).

Understanding these steps allows us to find inverses confidently by systematically undoing the actions applied in the original function.