Problem 55 - College Algebra with Modeling and Visualization

Question

Find a symbolic representation for \(f^{-1}(x).\) $$ f(x)=\frac{x}{x+2} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We are given a function \(f(x) = \frac{x}{x+2}\) and we need to find its inverse function, denoted by \(f^{-1}(x)\). This involves finding a function that, when composed with \(f\), returns the original input value.

Step 2: Replace f(x) with y

First, we write the function as \(y = \frac{x}{x+2}\). This step helps us find \(x\) in terms of \(y\), which is the key to finding the inverse function.

Step 3: Swap x and y

Next, interchange \(x\) and \(y\) in the equation to solve for the inverse: \(x = \frac{y}{y+2}\). This reflects the idea of reversing the function.

Step 4: Solve for y in terms of x

Multiply both sides of \(x = \frac{y}{y+2}\) by \((y+2)\) to eliminate the fraction: \(x(y+2) = y\). Then, expand and rearrange the equation: \(xy + 2x = y\). Isolate \(y\): \(xy - y = -2x\), or \((x-1)y = -2x\). Finally, solve for \(y\): \(y = \frac{-2x}{x-1}\).

Step 5: Write the Inverse Function

After finding \(y = \frac{-2x}{x-1}\), we identify this expression as the inverse function. Thus, \(f^{-1}(x) = \frac{-2x}{x-1}\).

Key Concepts

Function Composition

Function composition is a fundamental concept in mathematics, it involves applying one function to the results of another. When we compose two functions, we obtain a new function. This process is often written as \( (f \circ g)(x) = f(g(x)) \). The essence of function composition lies in input-output relations. Each function processes inputs in its own way, and when composed, the output of one function becomes the input to the next.

In the context of inverse functions, composition is a handy tool. If you have a function \(f\) and its inverse \(f^{-1}\), composing these as \((f \circ f^{-1})(x)\), or \( (f^{-1} \circ f)(x) \), should return \(x\). This property is what essentially defines two functions as inverses. It ensures that one function undoes the operation of the other, reverting to the original input.

Understanding function composition helps in verifying if functions are truly inverses of each other. When working with functions, always consider how they interact through composition to maintain the integrity of input-output transformations.

Solving Equations

Solving equations is a critical step in finding inverse functions. This algebraic process typically involves isolating the variable of interest by performing operations that 'undo' what's in the equation. In the discussed solution, you start with \(x = \frac{y}{y+2}\) and aim to solve for \(y\).

Let's break this down further:

• First, eliminate any fractions by multiplying both sides by the denominator.
• Then, expand the equation if necessary, which helps in collecting all terms involving the variable you want to solve for.
• Next, consolidate terms that contain the variable on one side of the equation.
• Finally, factor and divide to isolate the variable.

These steps are universal in algebra and are necessary to understand not only for inverse functions, but for solving all sorts of equations in mathematical problems.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to solve equations or make them more comprehensible. It's an essential skill when working to find inverse functions, especially when equations become complex.

When manipulating the equation \(xy + 2x = y\), the goal is to rearrange it such that \(y\) is isolated on one side:

• Start by shifting terms around to bring variable terms together.
• Look for common factors to simplify the expression.
• In this example, factor out \(y\) from both sides to expose common terms.
• To isolate \(y\), divide by the grouped coefficient, leaving \(y\) on one side of the equation.

Through effective algebraic manipulation, you can achieve a clearer expression of the inverse function \(y = \frac{-2x}{x-1}\). These principles of manipulation are staples in algebra, aiding in solving and understanding math expressions.