Problem 49 - Algebra and Trigonometry

Question

Find the inverse function of \(f\) $$ f(x)=\frac{1+3 x}{5-2 x} $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \frac{5x - 1}{3 + 2x}\).

Step 1: Replace f(x) with y

Start by replacing the function notation \(f(x)\) with \(y\). This means we write the equation as \(y = \frac{1+3x}{5-2x}\).

Step 2: Swap x and y

The next step in finding the inverse function is to swap \(x\) and \(y\). This gives us \(x = \frac{1+3y}{5-2y}\). Now, solve this equation for \(y\).

Step 3: Clear the fraction

Multiply both sides of the equation by the denominator \(5-2y\) to eliminate the fraction:\[ x(5 - 2y) = 1 + 3y \]

Step 4: Expand and Rearrange

Expand the left side of the equation and rearrange the terms:\[ 5x - 2xy = 1 + 3y \]Bring terms involving \(y\) to one side and constant terms to the other:\[ 5x - 1 = 3y + 2xy \]

Step 5: Factor out y

Factor \(y\) from the terms on the right side:\[ 5x - 1 = y(3 + 2x) \]

Step 6: Solve for y

Divide both sides by \(3 + 2x\) to solve for \(y\):\[ y = \frac{5x - 1}{3 + 2x} \]Thus, the inverse function is \(f^{-1}(x) = \frac{5x - 1}{3 + 2x}\).

Key Concepts

Algebraic Manipulation

When solving for an inverse function, algebraic manipulation is a key process. It's like solving a puzzle where the goal is to isolate a variable—in this case, the variable we swapped to become the new output.
Let's explore how this unfolds:

• First, we swap the roles of the dependent and independent variables. Our initial equation is given as a function of x. By swapping, x becomes dependent on y, and vice versa.
• Then, we rearrange the equation to solve for y. This involves moving terms around and potentially factoring out variables. Take a strategy like combining like terms or isolating y the same way you might expose one piece of a puzzle to manipulate it more easily.
• Techniques like expanding and simplifying expressions are crucial. We use distributive properties to expand, then collect terms involving y on one side to facilitate factoring and eventually isolating y.

Algebraic manipulation is about systematically applying operations like addition, subtraction, multiplication, and division to neatly isolate the desired variable. This is what allows us to reveal the inverse function.

Function Notation

Function notation gives us a structured way to represent transformations between variables. It's a mathematical language that signals how functions behave and relate to different inputs and outputs.
Here's a look at how this comes into play:

• We begin with the original function notation, f(x), which is simply a way of labeling the output of a function for a given input x. This allows for a clear depiction of mathematical relationships.
• To find an inverse, we represent the function in terms of y, known as the dependent variable. In our example, we swap f(x) with y to ease the process of finding the inverse.
• Once we find the inverse, we express it using the inverse function notation, written as \( f^{-1}(x) \). This notation clearly indicates the new rules of transformation from input back to output.

Function notation helps keep track of how each variable influences the other and simplifies complex formulas into understandable transformations.

Fraction Elimination

Dealing with fractions in an equation can complicate the process of finding inverses. Fraction elimination involves removing these fractions to simplify the equation.
Let's break it down:

• The first step is identifying fractions in the equation. Typically, these occur when the variable is in the numerator or denominator.
• We eliminate fractions by multiplying both sides of the equation by the denominator, effectively clearing the fraction. For instance, in the given problem, we multiply by \( 5-2y \) to clear the fraction.
• This multiplication must be applied across the entire equation, allowing us to shift our focus to simpler manipulations, like addition and subtraction. The aim is to arrive at an easier form to handle algebraically.

Eliminating fractions streamlines the equation, setting the stage for solving, rearranging, and ultimately isolating the desired variable.