Problem 22 - College Algebra

Question

The functions in Exercises \(11-30\) 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 $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$f(x)=\frac{2}{x}$$

Step-by-Step Solution

Answer

The inverse of the function \(f(x) = \frac{2}{x}\) is \(f^{-1}(x) = \frac{2}{x}\). Verification with the composite functions \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) confirms that this indeed the correct inverse.

Step 1: Derive the inverse function

First, we replace \(f(x)\) with \(y\) to get \(y = \frac{2}{x}\). Then we swap \(x\) and \(y\) to get \(x = \frac{2}{y}\). Solving this equation for \(y\) gives the inverse of our original function. Cross-multiplying gives \(xy = 2\), and solving for \(y\) gives \(y = \frac{2}{x}\). So, our inverse function is \(f^{-1}(x) = \frac{2}{x}\).

Step 2: Verify the first composite function

In order to confirm that the inverse function is correct, we use it in the composition \(f(f^{-1}(x))\). In place of \(f(x)\), we substitute our original function \(f(x) = \frac{2}{x}\) and for \(f^{-1}(x)\), we substitute our computed inverse \(f^{-1}(x) = \frac{2}{x}\). This gives us \(f(f^{-1}(x)) = \frac{2}{\frac{2}{x}}\). Cancelling out \(2\) in the numerator and denominator we get \(f(f^{-1}(x)) = x\).

Step 3: Verify the second composite function

Next, we check the composition \(f^{-1}(f(x))\). In this case, \(f(x)\) is our original function \(f(x) = \frac{2}{x}\) and \(f^{-1}(x)\) is our computed inverse \(f^{-1}(x) = \frac{2}{x}\). Therefore, \(f^{-1}(f(x)) = \frac{2}{\frac{2}{x}}\). Again, by cancelling out \(2\) in the numerator and denominator we get \(f^{-1}(f(x)) = x\). This completes the verification.

Key Concepts

Composition of Functions

When we talk about the composition of functions, we're essentially plugging one function into another. In this context, if you have two functions, say \( f(x) \) and \( g(x) \), the composition can take two forms: \( f(g(x)) \) and \( g(f(x)) \). For functions to have an inverse relationship, it is crucial that these compositions simplify to \( x \).
For example, in the exercise given, one composition \( f(f^{-1}(x)) \) effectively substitutes everything in the inverse function back into the original function. It looks like:

• Start with \( f^{-1}(x) = \frac{2}{x} \)
• Replace the \( x \) in \( f(x) = \frac{2}{x} \) with \( f^{-1}(x) \): \( f\left(f^{-1}(x)\right) = \frac{2}{\frac{2}{x}} \)
• This simplifies directly back to \( x \), proving the correctness of both equations.

Understanding this composition is key to recognizing whether two functions are indeed inverses of each other.

One-to-one Functions

A critical concept in finding inverse functions is understanding if the original function is one-to-one. A function is one-to-one if every output value is linked to exactly one input value. This means, for every \( y \), there is a single \( x \), and vice versa.
A simple test to determine if a function is one-to-one is the horizontal line test. If a horizontal line intersects the graph of the function at no more than one point, the function is one-to-one.
In terms of algebra, a one-to-one function assures that its inverse will also be a valid function. So in the exercise, since \( f(x) = \frac{2}{x} \) is a one-to-one function, we are guaranteed that its inverse \( f^{-1}(x) = \frac{2}{x} \) is also valid. This characteristic of one-to-one functions lets us rely on inverse symmetry when verifying compositions.

Verifying Inverse Functions

Verifying inverse functions is a method to ensure that two functions truly are inverses of each other. This involves two main steps: checking if \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
When you perform these checks, you should end up with \( x \) for both cases if the functions are truly inverses. Here's why:

• For \( f(f^{-1}(x)) \), you're using the inverse function to basically "undo" whatever \( f \) did, bringing you back to your starting \( x \).
  
• Similarly, \( f^{-1}(f(x)) \) starts with \( f \) and then applies \( f^{-1} \) to return to that initial value \( x \).

In our example, when we replaced \( x \) with \( f^{-1}(x) \) in \( f \), and vice versa, both compositions simplified perfectly to \( x \). This congruence shows that they successfully are inverse functions.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to isolate or solve for a specific variable. It is a fundamental skill when dealing with functions and inverses.
In the exercise, the algebraic steps were crucial:

• For deriving \( f^{-1}(x) \) from \( f(x) = \frac{2}{x} \), we first defined \( y = \frac{2}{x} \).
• Swapping \( x \) and \( y \) to find the inverse gives \( x = \frac{2}{y} \).
• Solving for \( y \) meant cross-multiplying to get \( xy = 2 \) and then isolating \( y \), resulting in \( y = \frac{2}{x} \).

These steps allow us to change the structure of an equation and find meaningful relationships between the variables. Practicing algebraic manipulation enhances the ability to tackle similar problems with confidence and precision.