Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=-\frac{x}{3}+1 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = -3x + 3 \). Verified calculations confirm that \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \).

Step 1: Begin with the function equation

The given function is \( f(x) = -\frac{x}{3} + 1 \). We need to find its inverse, so start by setting \( y = f(x) \), which gives us the equation \( y = -\frac{x}{3} + 1 \).

Step 2: Solve for x in terms of y

Rearrange the equation \( y = -\frac{x}{3} + 1 \) to express \( x \) in terms of \( y \). Subtract 1 from both sides: \( y - 1 = -\frac{x}{3} \). Multiply both sides by -3: \( x = -3(y - 1) \). Thus, \( x = -3y + 3 \).

Step 3: Replace y with x to find the inverse function

Now that we have \( x = -3y + 3 \), swap \( x \) and \( y \) to find the inverse function. This gives us \( y = -3x + 3 \), which means the inverse function is \( f^{-1}(x) = -3x + 3 \).

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

Substitute \( f(x) \) into the inverse function: \( f^{-1}(f(x)) = f^{-1}(-\frac{x}{3} + 1) \). Substitute \( u = -\frac{x}{3} + 1 \) into the inverse function: \( -3u + 3 = -3(-\frac{x}{3} + 1) + 3 \). Simplifying gives \( x \), thereby confirming that \( f^{-1}(f(x)) = x \).

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

Substitute \( f^{-1}(x) \) into the original function: \( f(f^{-1}(x)) = f(-3x + 3) \). Compute this as \( -\frac{(-3x + 3)}{3} + 1 \). Simplifying gives \( x \), thereby confirming that \( f(f^{-1}(x)) = x \).

Key Concepts

Function Composition

Function composition is the process of combining two functions to create a new function. It's like applying one function and then applying another to the result. In the context of inverse functions, we often use composition to check correctness. By composing a function and its inverse, we should end up back with the original input, symbolized by the equation:

• \( f^{-1}(f(x)) = x \)
• \( f(f^{-1}(x)) = x \)

Function composition is essential because it validates that the inverse function genuinely 'undoes' the effect of the original function. When you plug the output of one function back into its inverse (or vice versa), the composition should return the starting variable, which is "\( x \)" in this situation. This property ensures the correct establishment of inverse relationships between two functions.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to isolate variables and solve problems. It involves a series of operations - like addition, subtraction, multiplication, and division - applied judiciously to both sides of an equation.To find an inverse function, we typically use algebraic manipulation to solve for one variable in terms of another. Starting with the original function \( f(x) = -\frac{x}{3} + 1 \), we replace \( f(x) \) with \( y \), leading to the equation:

• \( y = -\frac{x}{3} + 1 \)

The goal is to express \( x \) as a function of \( y \). By performing algebraic steps - subtracting 1 from \( y \) and then multiplying throughout by -3 - we isolate \( x \). This process reveals the inverse function, \( f^{-1}(x) = -3x + 3 \). Mastering these manipulative techniques is key to confidently dealing with inverse functions in algebra.

Verification of Inverse Functions

Verification of inverse functions is crucial in mathematics. This step ensures that the computed inverse actually behaves as expected. To verify an inverse function, we make use of function composition, confirming both \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) result in \( x \).For our problem, this means we need to:

• Insert \( f(x) \) into \( f^{-1} \): Start with \( f^{-1}(f(x)) \). Replacing \( f(x) \, (=-\frac{x}{3}+1) \) back into \( f^{-1}\), simplifies back to \( x \).
• Insert \( f^{-1}(x) \) into \( f \): In the form \( f(f^{-1}(x)) \). Doing this steps through simplifying \( -\frac{(-3x + 3)}{3} + 1 \), which also results in \( x \).

These verifications prove that our original function and its inverse function are mutually accurate inverses, always returning the original input when composed with each other. This confirmation is indispensable for being sure about our calculations and supports the correctness of the inverse relationships we establish.