Problem 29 - Intermediate Algebra

Question

Find the inverse of the given function by using the "undoing process," and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). (Objective 4) $$f(x)=\frac{1}{3} x-\frac{2}{5}$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = 3x + \frac{6}{5} \); both compositions verify the inverse.

Step 1: Identify the Function

The given function is \( f(x) = \frac{1}{3}x - \frac{2}{5} \). We are going to find its inverse by using the undoing process.

Step 2: Replace the Function Notation with a Variable

Replace \( f(x) \) with \( y \). The equation becomes \( y = \frac{1}{3}x - \frac{2}{5} \).

Step 3: Solve for x in terms of y

Our goal is to express \( x \) in terms of \( y \). Begin by isolating \( x \) on one side of the equation:1. Add \( \frac{2}{5} \) to both sides: \( y + \frac{2}{5} = \frac{1}{3}x \).2. Multiply both sides by 3 to isolate \( x \): \( 3y + \frac{6}{5} = x \).Hence, \( x = 3y + \frac{6}{5} \).

Step 4: Swap Variables to Find the Inverse

Now, replace \( y \) with \( x \) to write the inverse function: \( f^{-1}(x) = 3x + \frac{6}{5} \).

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

Verify by composing \( f \) and \( f^{-1} \):\[ f(f^{-1}(x)) = f(3x + \frac{6}{5}) = \frac{1}{3}(3x + \frac{6}{5}) - \frac{2}{5} = x + \frac{2}{5} - \frac{2}{5} = x \].Therefore, \(\left(f \circ f^{-1}\right)(x) = x\).

Step 6: Verify \(\left(f^{-1} \circ f\right)(x) = x\)

Verify by composing \( f^{-1} \) and \( f \):\[ f^{-1}(f(x)) = f^{-1}(\frac{1}{3}x - \frac{2}{5}) = 3(\frac{1}{3}x - \frac{2}{5}) + \frac{6}{5} = x - 2 + \frac{6}{5} = x \].Thus, \(\left(f^{-1} \circ f\right)(x) = x\).

Key Concepts

Function Composition

Function composition is like stacking two machines—output from one becomes input for the next. In mathematics, we use the composition of functions to combine them. For any functions \( f \) and \( g \), their composition, denoted \( f \circ g \), means we first apply \( g \) then \( f \). For inverse functions specifically, composition verifies the correctness. When we compose a function \( f \) with its inverse \( f^{-1} \), we expect:

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

This shows that applying a function and then its inverse does nothing, meaning both operations undo each other.In practice, using composition to verify inverse functions ensures they properly "cancel out" their effects, returning to the original input.

Linear Functions

A linear function creates a straight line when graphed and has a general form of \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. The function \( f(x) = \frac{1}{3}x - \frac{2}{5} \) is linear, with:

• Slope (\( m \)): \( \frac{1}{3} \)
• Y-intercept (\( b \)): \(-\frac{2}{5} \)

Linear functions are straightforward for calculating inverses because their simplicity allows direct operations on the formula. When finding the inverse of this linear function, you swap the input and output, solving for the original input in terms of the output. This mathematical process reconfigures the function’s operations, and practically, you "reverse" its effect.

Function Verification

Function verification involves checking the correctness of an inverse by composition. After finding an inverse function, performing verification confirms it functions as intended. This is done by utilizing both:

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

By substituting the expressions into these compositions, simplifying reveals whether the end result is indeed the original input \( x \). In the exercise example, through painstaking verification:

• \( f(3x + \frac{6}{5}) = x \)
• \( f^{-1}(\frac{1}{3}x - \frac{2}{5}) = x \)

Both return \( x \), illustrating the inverse function accurately reverses \( f \), confirming correctness through logical steps.