Problem 83 - Elementary and Intermediate Algebra

Question

Find the inverse of \(f(x)=\frac{x+1}{x-1}\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the function

We are given a function \( f(x) = \frac{x+1}{x-1} \) and we need to find its inverse. The inverse function, denoted \( f^{-1}(x) \), will be such that \( f(f^{-1}(x)) = x \). Our aim is to revert the roles of the inputs and outputs of the function.

Step 2: Rewriting the equation

Start by replacing \( f(x) \) with \( y \) to make it easier to work with: \( y = \frac{x+1}{x-1} \). Our task is to solve this equation for \( x \) in terms of \( y \).

Step 3: Cross-multiplying to eliminate the fraction

Eliminate the fraction by multiplying both sides by \( (x-1) \), which gives us: \( y(x-1) = x + 1 \).

Step 4: Distribute and rearrange terms

Distribute \( y \) on the left-hand side: \( yx - y = x + 1 \). Rearrange this equation to collect all terms involving \( x \) on one side. This results in: \( yx - x = y + 1 \).

Step 5: Factor out the common term

Factor out \( x \) from the terms on the left: \( x(y - 1) = y + 1 \). This step helps to isolate \( x \).

Step 6: Solve for x

To isolate \( x \), divide both sides by \( y - 1 \): \( x = \frac{y + 1}{y - 1} \). This equation represents \( x \) in terms of \( y \), which is actually the inverse function.

Step 7: Substitute back to find the inverse

Since \( x = \frac{y + 1}{y - 1} \) is in terms of \( y \), replace \( y \) with \( x \) to get the inverse function: \( f^{-1}(x) = \frac{x + 1}{x - 1} \).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a powerful tool used to rearrange and simplify equations. It involves performing operations like addition, subtraction, multiplication, and division to isolate variables or express equations in different forms. In the process of finding the inverse function, algebraic manipulation plays a critical role. Our main goal is to isolate one variable in terms of another by rearranging terms.

• Step 1: Initially, we have the equation for a function given as \( y = \frac{x+1}{x-1} \).
• Step 2: We need to express \( x \) in terms of \( y \). This requires careful manipulation and rearranging terms.

Understanding how to shift terms around and employ basic algebraic operations helps to progress through problems efficiently. Algebraic manipulation is essentially the backbone of transforming the function and finding its inverse.

Solving Equations

Solving equations involves finding the values of an unknown that make the equation true. When finding an inverse function, solving the equation is necessary to express one variable in terms of another. Once we have the equation \( y(x-1) = x + 1 \), solving it takes systematic steps:

• Distribute \( y \) across \( (x-1) \) to get \( yx - y = x + 1 \).
• Rearrange the equation so that all terms involving \( x \) are on one side: \( yx - x = y + 1 \).
• Factor out the common term \( x \), which gives \( x(y-1) = y + 1 \).

Each step aims to simplify the equation and gather like terms to eventually isolate \( x \). Solving these equations requires strategic thinking and a clear plan for each manipulation.

Cross-Multiplication

Cross-multiplication is a method used to eliminate fractions and is particularly useful in solving rational equations. In our exercise, it helps simplify the equation so we can isolate the desired variables. Given \( y = \frac{x+1}{x-1} \), cross-multiplication involves multiplying both sides by \( (x-1) \):

• This step effectively removes the denominator yielding the equation \( y(x-1) = x + 1 \).
• Once the fraction is gone, the equation becomes easier to handle by traditional algebraic methods.

Employing cross-multiplication simplifies the initial setup of equations, making it more straightforward to proceed with additional algebraic operations. It is instrumental in transitioning from a complex equation with fractions to a simpler linear representation, which paves the way for further manipulation.