Problem 28 - Precalculus Essentials

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 \(f(x) = \frac{2x - 3}{x + 1}\) is \(f^{-1}(x) = \frac{x + 3}{2 - x}\)

Step 1: Express x in terms of y

To find the inverse of a function, the first step is to replace the function notation \(f(x)\) with the variable \(y\) for easier manipulation. The equation now reads \(y = \frac{2x - 3}{x + 1}\)

Step 2: Interchange x and y

By switching \(x\) and \(y\) in the equation, we establish the inverse relationship. Now the equation becomes \(x = \frac{2y - 3}{y + 1}\)

Step 3: Solve for y

We rearrange this equation to isolate \(y\) on one side (the form \(y = f^{-1}(x)\)). To do this, we cross multiply our equation to get rid of the denominator and then isolate y. This gives us \(y = f^{-1}(x) = \frac{x + 3}{2 - x}\)

Step 4: Verification

We must verify that \(f^{-1}(x)\) is indeed the inverse of \(f(x)\). Plugging \(f^{-1}(x)\) into \(f(x)\) should yield \(x\), as well as plugging \(f(x)\) into \(f^{-1}(x)\). We have \(f(f^{-1}(x)) = f\left(\frac{x + 3}{2 - x}\right) = x\) and \(f^{-1}(f(x)) = f^{-1}\left(\frac{2x - 3}{x + 1}\right) = x\). Therefore, we have verified that \(y = f^{-1}(x) = \frac{x + 3}{2 - x}\) is indeed the inverse of \(f(x) = \frac{2x - 3}{x + 1}\)

Key Concepts

Understanding One-to-One Functions

One-to-one functions, or injective functions, are fundamental in the study of mathematical functions and their inverses. In simplest terms, a function is called one-to-one when each input value maps to exactly one unique output value, and no two different inputs have the same output. This uniqueness is crucial when finding inverses of functions because it ensures that the inverse will also be a function - each output from the original function becomes an input for the inverse, leading to a single output.

To visualize a one-to-one function, you can use the 'horizontal line test'. If any horizontal line intersects the graph of the function at no more than one point, then the function is one-to-one. This test is a practical way to quickly assess if a function has an inverse that is also a function.

Finding Inverses of Functions

The process of finding an inverse function can be seen as flipping a function over the line where y equals x. This is because the inverse essentially switches each input with its corresponding output. To find an inverse function, typically you follow these crucial steps:

Step 1: Swap Variables

Replace the function notation with 'y' to work with a familiar 'y = ...' format, and then interchange the x and y variables. This step sets up the equation to solve for the new 'y', which is the inverse function.

Step 2: Solve for New y

After swapping the variables, rearrange the equation to solve for the new y (which is actually the inverse function, denoted as f^{-1}(x)). This often involves algebraic manipulation such as cross multiplication to isolate y.

Step 3: Verify

Always verify your proposed inverse by checking that both f(f^{-1}(x)) and f^{-1}(f(x)) return the original input x. This confirms that the function and its inverse are correctly paired.

Comprehending Function Composition

Function composition involves applying one function to the results of another function. When we denote composition, we use the notation \( f \circ g \) which means f(g(x)). The fundamental purpose behind composition is to understand how two functions combine to produce a new function. This concept is especially relevant when dealing with inverses - the composition of a function and its inverse yields the identity function, which simply returns the original input value.

For function inverses, we focus on the composition \( f(f^{-1}(x)) \) and its counterpart \( f^{-1}(f(x)) \), both of which should result in x if f is one-to-one and has a proper inverse. This composition is a powerful tool in verifying the correctness of an inverse function, as showcased in the exercise solution provided.

The elegance of function composition lies in its symmetry when dealing with inverses: no matter which function is applied first, if one is the inverse of the other, the result will always 'undo' the effect of the first and leave you with the initial value.