Problem 28 - College Algebra

Question

The functions are all one-to-one. For each function, a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$ f(x)=\frac{2 x-3}{x+1} $$

Step-by-Step Solution

Answer

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

Step 1: Find the inverse function

The given function is \(f(x) = \frac{2x - 3}{x+1}\). To find the inverse function, replace \(f(x)\) with \(y\), which looks like \(y = \frac{2x - 3}{x+1}\), then swap \(x\) and \(y\) to get \(x = \frac{2y - 3}{y+1}\). Next, solve the resulting equation for \(y\) to obtain the inverse function. Cross-multiply to get \(x(y+1) = 2y - 3\), then distribute the \(x\) to get \(xy + x = 2y - 3\). Gather the \(y\)-terms on one side and the \(x\)-terms on the other to get \(2y - xy = x + 3\), then factor out \(y\) to get \(y(2 - x) = x + 3\), finally isolate \(y\) to get \(y = \frac{x + 3}{2 - x}\), which is \(f^{-1}(x)\).

Step 2: Verify the first property for inverse functions

To verify that \(f(f^{-1}(x)) = x\), plug \(f^{-1}(x)\) into \(f(x)\), it looks like \(f(f^{-1}(x)) = f\left(\frac{x + 3}{2 - x}\right) = \frac{2\left(\frac{x + 3}{2 - x}\right) - 3}{\frac{x + 3}{2 - x} + 1} = \frac{2x + 6 - 3(2 - x)}{x + 3 + 2 - x} = \frac{2x - 3}{x+1} = x\). So it is confirmed that \(f(f^{-1}(x)) = x\).

Step 3: Verify the second property for inverse functions

To verify that \(f^{-1}(f(x)) = x\), plug \(f(x)\) into \(f^{-1}(x)\), it looks like \(f^{-1}(f(x)) = f^{-1}\left(\frac{2x - 3}{x + 1}\right) = \frac{\frac{2x - 3}{x + 1} + 3}{2 - \frac{2x - 3}{x + 1}} = \frac{2x - 3 + 3(x + 1)}{2x - 3 + x + 1} = \frac{x(2 - x)}{x(2 - x)} = x\). So it is confirmed that \(f^{-1}(f(x)) = x\).

Key Concepts

One-to-One Functions

One-to-one functions play a crucial role when discussing inverse functions. A function is called one-to-one if every element of the function's domain maps to a unique element in its range. More clearly, for any two distinct elements,

• If \(f(a) = f(b)\), then it must be that \(a = b\).
• This ensures that no horizontal line intersects the graph of the function at more than one point.

The property of being one-to-one is essential for a function to have an inverse that is also a function. If a function is one-to-one, we can find an inverse function by swapping the input and output throughout. This step is crucial because not all functions have inverses.
A quick test for one-to-one authenticity is the Horizontal Line Test: if any horizontal line crosses the graph of the function more than once,

• it indicates that the function is not one-to-one, and thus doesn't have an inverse that is a function.
• Our function \(f(x) = \frac{2x - 3}{x+1}\) is one-to-one, so it qualifies for having an inverse function.

Verification of Inverse Functions

Verification of inverse functions ensures that the calculated inverse is correct. To verify an inverse function \(f^{-1}(x)\), we need to check two conditions:

• First, substituting \(f^{-1}(x)\) into \(f(x)\) should result in \(x\).
• Second, substituting \(f(x)\) into \(f^{-1}(x)\) should also give back \(x\).

This means that the composition of a function and its inverse in either order should return the original input value. For our example:

• We have \(f(f^{-1}(x)) = f\left(\frac{x + 3}{2 - x}\right)\) which simplifies to \(x\), showing validity.
• Likewise, \(f^{-1}(f(x)) = f^{-1}\left(\frac{2x - 3}{x+1}\right)\) simplifies to \(x\), confirming correctness.

These verifications help confirm that the computed inverse function is accurate, maintaining that every input and output pair in the original function is reversed in the inverse.

Algebraic Manipulation

Algebraic manipulation is a systematic way of arranging equations to make a certain goal easier accomplished, such as finding an inverse function. It involves steps such as solving for variables by rearranging them, distributing, and factoring.

• First, for the given function \(f(x) = \frac{2x - 3}{x+1}\), we start by replacing \(f(x)\) with \(y\), to get \(y = \frac{2x - 3}{x+1}\).
• Next, swap \(x\) and \(y\) to begin solving for \(y\).
• This gives us the equation \(x = \frac{2y - 3}{y+1}\). Cross-multiply to eliminate the fraction, leading to \(x(y+1) = 2y - 3\).
• Distribute, then combine like terms, and factor, arriving at \(y = \frac{x + 3}{2 - x}\).

By working through these algebraic steps carefully and methodically, we derive the inverse function correctly. Paying attention to these processes ensures clarity and correctness in finding inverse functions and engaging with other complex algebraic problems.