Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(31-44\), find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\). 31\. \(f(x)=x+1\)

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = x - 1\). Both verifications hold true.

Step 1: Identify the function

The given function is \( f(x) = x + 1 \). We need to find its inverse function, which we denote as \( f^{-1}(x) \).

Step 2: Set up the equation to find the inverse

To find the inverse function, start by setting \( y = f(x) = x + 1 \). Now, our goal is to express \( x \) in terms of \( y \).

Step 3: Solve for the inverse

Rearrange the equation \( y = x + 1 \) to solve for \( x \). This gives us \( x = y - 1 \). So, the inverse function is \( f^{-1}(x) = x - 1 \).

Step 4: Verify \( f^{-1}(f(x)) = x \)

Substitute \( f(x) = x + 1 \) into the inverse function: \( f^{-1}(f(x)) = f^{-1}(x + 1) = (x + 1) - 1 = x \). This confirms that \( f^{-1}(f(x)) = x \).

Step 5: Verify \( f(f^{-1}(x)) = x \)

Substitute \( f^{-1}(x) = x - 1 \) into the original function: \( f(f^{-1}(x)) = f(x - 1) = (x - 1) + 1 = x \). This confirms that \( f(f^{-1}(x)) = x \).

Key Concepts

Function Composition

Function composition involves putting one function inside another. When dealing with functions, we often find it useful to connect two functions into a single operation. For two functions, say \( f(x) \) and \( g(x) \), the composition is denoted as \( (f \circ g)(x) \) or \( f(g(x)) \). In essence, you take the output from the inner function \( g(x) \) and input it into the outer function \( f(x) \).

For example, let's take the original function \( f(x) = x + 1 \). If you have another function \( g(x) \), you can create a new function by composing \( f(x) \) with \( g(x) \). Say \( g(x) = x^2 \), then their composition would be \( f(g(x)) = f(x^2) = x^2 + 1 \).

This is an important concept because it allows us to see how functions interact with one another. It lays the foundation for finding inverse functions, where a function's output reverts back to the original input.

Verification of Inverse Functions

When you find an inverse function, the next task is to verify it's correct. This involves proving two key equations: \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \). These equations confirm that a function and its inverse cancel each other out, leaving the original input value.

Let's look at the example \( f(x) = x + 1 \). We found that its inverse is \( f^{-1}(x) = x - 1 \). To verify, substitute \( f(x) \) into \( f^{-1}(x) \):
\[ f^{-1}(f(x)) = f^{-1}(x + 1) = (x + 1) - 1 = x.\]
This shows that the equation \( f^{-1}(f(x)) = x \) holds.

Next, replace \( f^{-1}(x) \) back into \( f(x) \):
\[ f(f^{-1}(x)) = f(x - 1) = (x - 1) + 1 = x.\]
This confirms \( f(f^{-1}(x)) = x \), affirming that the inverse function was correctly identified. This verification is critical in mathematics as it ensures that our understanding and calculations of inverse functions are precise and accurate.

Step-by-Step Solution

Breaking down a problem into clear, sequential steps can simplify the process of finding an inverse function and verifying it. Let's take the function \( f(x) = x + 1 \), and illustrate each step involved in handling it step-by-step.

1. **Identify the function**:
- Start with the given function: \( f(x) = x + 1 \). Determine it's a linear function.

2. **Set up to find the inverse**:
- Assign \( y \) to \( f(x) \), giving \( y = x + 1 \). Your task here is to solve this for \( x \) to express \( f^{-1}(x) \).

3. **Solve for the inverse**:
- Rearrange \( y = x + 1 \) to isolate \( x \): \( x = y - 1 \), showing \( f^{-1}(x) = x - 1 \).

4. **Verify inverse using composition**:
- Substitute \( f(x) \) into \( f^{-1}(x) \) and check \( f^{-1}(f(x)) = x \).
- Replace \( f^{-1}(x) \) into \( f(x) \) and ensure \( f(f^{-1}(x)) = x \).

By following these steps, you break the problem into manageable parts, making it easier to understand and verify. This process ensures your solution is robust and accurate, which is vital in mathematical problem solving.