Problem 22 - College Algebra

Question

For the following exercises, find the inverse of the functions. $$ f(x)=\frac{2}{x+8} $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{2 - 8x}{x} \).

Step 1: Replace f(x) with y

Begin by letting the function be written as \( y = \frac{2}{x+8} \). This helps in equating to find the inverse later.

Step 2: Swap x and y

To find the inverse, swap \( x \) and \( y \) in the equation: \( x = \frac{2}{y+8} \). This transformation reflects the inverse relationship.

Step 3: Solve for y

Now, manipulate the equation to solve for \( y \). First, multiply both sides by \( y + 8 \) to eliminate the fraction: \( x(y+8) = 2 \).

Step 4: Distribute x and Simplify

Distribute \( x \) into \( y+8 \): \( xy + 8x = 2 \). Then isolate \( y \) by moving \( 8x \) to the other side: \( xy = 2 - 8x \).

Step 5: Isolate y

Finally, solve for \( y \) by dividing both sides by \( x \): \( y = \frac{2 - 8x}{x} \). This gives the equation of the inverse function.

Step 6: Write the inverse function

Change the variable back from \( y \) to \( f^{-1}(x) \) to denote the inverse of the original function: \( f^{-1}(x) = \frac{2 - 8x}{x} \).

Key Concepts

Function Notation

Function notation simply refers to how we represent functions in a mathematical form. In this exercise, you encounter the function notation \( f(x) \), which is read as "\( f \) of \( x\)." This notation helps us identify that \( f \) is a function concerning \( x \).
When finding inverse functions, understanding function notation is crucial since we essentially reverse the roles of input and output values. For example, replacing \( f(x) \) with \( y \) makes it easier to manipulate the equation for inverse calculations. Switching to \( f^{-1}(x) \) later explicitly denotes the inverse function. This conveys that we have distinguished this separate function in relation to the original \( f(x) \).

• Using \( y \) instead of \( f(x) \) clarifies algebraic manipulations.
• The inverse function \( f^{-1}(x) \) indicates the reverse operation.

Algebraic Manipulation

Algebraic manipulation involves applying mathematical operations to formulas or equations to derive a solution. When finding the inverse of a function, algebraic manipulation is necessary to isolate terms, simplify expressions, and solve for unknown variables.
In the provided solution, we first swapped \( x \) and \( y \) to begin the process of uncovering the inverse. Then, various algebraic strategies were employed to solve for \( y \), the inverse function's independent variable. Here’s a breakdown of it:

• Swapping \( x \) and \( y \): This step lays the foundation for finding the inverse.
• Eliminating fractions: Multiplying both sides by \( y + 8 \) helps remove the fraction, making it simpler to manipulate.
• Isolating \( y \): Use distributive law and basic algebra to isolate \( y \).

These strategies are valuable tools in algebra that simplify complex identities by transforming them into actionable equations.

Fractions in Equations

Working with fractions in equations often seems daunting, but it's integral to mastering algebra. In this problem, fractions appear as part of the function's representation. Here's how you can tackle fractions to solve for the inverse function.
When encountering an equation like \( x = \frac{2}{y+8} \), your first step is to eliminate the fraction to handle the equation more easily. You achieve this by multiplying both sides by the denominator, \( (y+8) \), which gives \( x(y+8) = 2 \). This maneuver makes it possible to handle fractions effectively in algebraic expressions.

• Multiply both sides: Reduces complexity, making further manipulation easier.
• Distributive property: Helps expand expressions, moving away from fraction complexities.

Understanding these techniques allows you to manipulate equations effectively and solve not just for inverse functions but a wide variety of problems involving fractions.