Problem 55 - College Algebra with Corequisite Support

Question

For the following exercises, find the inverse of the functions with \(a, b, c\) positive real numbers. $$ f(x)=\frac{a x+b}{x+c} $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{b - xc}{x - a} \).

Step 1: Replace f(x) with y

Start by substituting \( y \) for \( f(x) \). We have: \[ y = \frac{ax + b}{x + c} \] Now, our task is to solve this equation for \( x \) to find the inverse of the function.

Step 2: Cross-Multiply to Clear Fraction

Multiply both sides of the equation by \( x + c \) to eliminate the fraction:\[ y(x + c) = ax + b \] This expansion will help us isolate \( x \) on one side.

Step 3: Distribute and Rearrange

Distribute \( y \) on the left-hand side of the equation:\[ yx + yc = ax + b \] Next, move all terms involving \( x \) to one side and constants to the other:\[ yx - ax = b - yc \]

Step 4: Factor Out x

Factor out \( x \) from the left side of the equation:\[ x(y - a) = b - yc \] This step isolates \( x \) with the remaining factors.

Step 5: Solve for x

Divide both sides by \( (y - a) \) to isolate \( x \):\[ x = \frac{b - yc}{y - a} \] Now this expression represents \( x \) in terms of \( y \), which is the inverse function.

Step 6: Write the Inverse Function

To express the inverse function, replace \( y \) with \( x \), since we are now reversing the roles. The inverse function is:\[ f^{-1}(x) = \frac{b - xc}{x - a} \]

Key Concepts

Function Notation

Function notation is a way to express mathematical situations using symbols and it provides an efficient way to describe mathematical operations. In function notation, a function named \( f \) with input \( x \) is denoted as \( f(x) \).
This tells us that the function \( f \) takes \( x \) as an input and maps it to an output value. In this exercise, the function \( f(x) = \frac{ax + b}{x + c} \) uses algebraic expressions to define that mapping explicitly.

Function notation is crucial when dealing with inverse functions. The notation \( f^{-1}(x) \) denotes the inverse of \( f(x) \).
This means if \( f(x) \) converts an input \( x \) to an output \( y \), then \( f^{-1}(y) \) will convert \( y \) back into \( x \). Using this notation helps us visualize and solve problems involving transformations and reversals of functions.

Cross-Multiplication

Cross-multiplication is a mathematical technique used to eliminate fractions from an equation, making the equation easier to solve. In the given exercise, we were tasked with finding the inverse function so we must clear the fractions caused by the original function definition.

In step 2 of the solution, cross-multiplication is used. We multiply both sides of the equation \( y = \frac{ax + b}{x + c} \) by \( x + c \) to eliminate the denominator. As a result, we obtain a simpler equation:

• \( y(x + c) = ax + b \)

Cross-multiplying by \( x + c \) effectively removes the fraction, enabling easier manipulation of the remaining terms.

Solving Equations

Solving equations involves finding the values for which a given equation holds true. When dealing with inverse functions, our goal is to express one variable in terms of another. In this exercise, we worked to solve for \( x \) in terms of \( y \).

After cross-multiplying, the equation becomes \( yx + yc = ax + b \). By arranging this equation, we can bring all terms with \( x \) on one side:

• \( yx - ax = b - yc \)

This process allows us to isolate \( x \) from the rest, leading to a logical path for solving for \( x \) each time we deal with inverse functions.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions through operations such as adding, subtracting, multiplying, and dividing terms.
This skill is invaluable when seeking the inverse of a function like \( f(x) = \frac{ax + b}{x + c} \). In the presented exercise, we use algebraic manipulation seriously to isolate \( x \) on one side of the equation.

From \( yx - ax = b - yc \), we factor out \( x \) on the left side:

• \( x(y-a) = b - yc \)

Then, by dividing both sides by \( (y-a) \), we succeed in finding \( x \):

• \( x = \frac{b - yc}{y - a} \)

Finally, replacing \( y \) with \( x \) gives us the inverse function. Such algebraic manipulation is a powerful tool in mathematics, routinely used to transform and simplify expressions as needed.