Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let $$ f(x)=\frac{2 x-1}{3 x+4} $$ Find \(f^{-1}(x)\)

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \frac{-4x-1}{3x-2}\).

Step 1: Substitute

Replace \(f(x)\) with \(y\). This gives us the equation \(y=\frac{2x-1}{3x+4}\).

Step 2: Swap \(x\) and \(y\)

Everywhere on the right side where there is an \(x\), replace it with \(y\), and everywhere where there is a \(y\), replace it with \(x\). This gives us \(x=\frac{2y-1}{3y+4}\).

Step 3: Solve for \(y\)

This is a somewhat complicated algebraic equation, but by proper cross-multiplication and rearrangement, we can isolate \(y\) on one side. First multiply both sides by \(3y+4\) which gives \(x(3y+4)=2y-1\). Then distribute \(x\) on the left side, giving \(3xy + 4x = 2y - 1\). Then rearrange the terms to get \(3xy - 2y = -4x -1\), and from this equation, we can factor out \(y\), to get \(y(3x - 2) = -4x -1\). Finally, divide both sides by \(3x-2\), giving a final answer of \(y=\frac{-4x-1}{3x-2}\). This expression for \(y\) is the inverse function.

Key Concepts

Functions

Functions are an essential building block in mathematics. They are like machines that take an input, do something with it, and produce an output. Formally, a function is a relation between a set of inputs (often called domain) and a set of possible outputs (called codomain) where each input is related to exactly one output.
In our case, the given function is \(f(x)=\frac{2x-1}{3x+4}\). Here:

• \(2x-1\) represents the operation applied to the input \(x\).
• \(3x+4\) in the denominator ensures the function behaves in a certain manner. It creates restrictions, for example, \(3x+4\) should not be zero to prevent division by zero.

Finding an inverse function \(f^{-1}(x)\) essentially means reversing the roles of inputs and outputs. If \(f\) takes \(x\) to \(y\), then \(f^{-1}\) takes \(y\) back to \(x\). This reverse setup helps us to understand the behavior of \(f\) in a comprehensive way.

Algebraic Manipulation

Algebraic manipulation involves transforming equations and expressions to make them simpler or to solve them. It involves using techniques and rules like distributing, factoring, and rearranging terms. In finding inverse functions, algebraic manipulation is crucial.
When we substitute \(f(x)\) with \(y\) in \(y=\frac{2x-1}{3x+4}\), and swap \(x\) and \(y\) to get \(x=\frac{2y-1}{3y+4}\), we are beginning the algebraic manipulation process. Solving for \(y\) involves a sequence of steps:

• Distributing: When \(x(3y+4)\) expands, it becomes \(3xy + 4x\).
• Rearranging terms: Bringing all \(y\)-related terms to one side and constants to the other helps in isolating \(y\).
• Factoring: Taking out the common terms, like \(y\), from the expression.
• Dividing: Simplifying the expression by dividing appropriate terms.

Each step requires careful application of algebraic rules to ensure the equation remains balanced and true.

Cross-Multiplication

Cross-multiplication is a convenient technique used to solve equations that involve fractions. It allows you to remove the fractions from an equation by multiplying each side of the equation by the denominators.
In our exercise, once we have the equation \(x = \frac{2y-1}{3y+4}\), cross-multiplication is used to eliminate the fraction. We multiply both sides by \(3y+4\), which results in:

• \(x(3y+4) = 2y - 1\). This multiplication effectively "gets rid of" the fraction, transforming the equation into one that is easier to manipulate algebraically.

From there, it’s about simplifying the resulting expression through algebraic manipulation, as described in the previous section. Cross-multiplication serves as a critical first step, allowing you to move forward with algebraic processes smoothly.