Problem 30 - 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) $$f(x)=\frac{2}{5} x+\frac{1}{3}$$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{5}{2}x - \frac{5}{6} \). Its composition with \( f \) in both orders returns \( x \).

Step 1: Identify Operations in the Function

The function given is \( f(x) = \frac{2}{5} x + \frac{1}{3} \). The operations applied to \( x \) are: first multiply by \( \frac{2}{5} \) and then add \( \frac{1}{3} \).

Step 2: Reverse the Operations

To find the inverse, we reverse the operations. Start by undoing the addition, then the multiplication.1. Subtract \( \frac{1}{3} \) from both sides: \( y - \frac{1}{3} = \frac{2}{5} x \).2. Multiply by the reciprocal of \( \frac{2}{5} \), which is \( \frac{5}{2} \):\[ x = \frac{5}{2} \left( y - \frac{1}{3} \right) \].

Step 3: Solve for the Inverse Function

Express \( x \) in terms of \( y \) to find the inverse: \[ x = \frac{5}{2}(y - \frac{1}{3}) \]Simplify it:\[ x = \frac{5}{2}y - \frac{5}{6} \]Therefore, the inverse function \( f^{-1}(x) = \frac{5}{2}x - \frac{5}{6} \).

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

Substitute \( f^{-1}(x) \) in \( f \):\[ f(f^{-1}(x)) = f\left( \frac{5}{2}x - \frac{5}{6} \right) \]Use the expression of \( f \):\[ = \frac{2}{5}\left( \frac{5}{2}x - \frac{5}{6} \right) + \frac{1}{3} \]Simplify inside the brackets:\[ = x - \frac{1}{3} + \frac{1}{3} \]Result:\[ = x \].

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

Substitute \( f(x) \) in \( f^{-1} \):\[ f^{-1}(f(x)) = f^{-1}\left( \frac{2}{5}x + \frac{1}{3} \right) \]Use the expression of \( f^{-1} \):\[ = \frac{5}{2}\left( \frac{2}{5}x + \frac{1}{3} - \frac{1}{3} \right) \]Simplify inside:\[ = \frac{5}{2} \times \frac{2}{5}x \]Result:\[ = x \].

Key Concepts

Functional Composition

One of the fundamental concepts in mathematics is functional composition, which is like plugging one function into another. This concept is essential when working with inverse functions. When you hear about composition, imagine solving a layered puzzle. You start from the inner piece and work your way outwards.

For example, if you have two functions, let's say \( f(x) \) and \( g(x) \), the composition \((f \circ g)(x)\) means you first apply \( g \) to \( x \) and then apply \( f \) to the result. It's like a two-step journey where the outcome of the first step is the starting point of the second.

• Begins with operation: Take input \( x \).
• Apply \( g \): Compute \( g(x) \).
• Apply \( f \): Compute \( f(g(x)) \).

This simple but powerful idea allows us to verify inverse functions. Specifically, for functions and their inverses \( f \) and \( f^{-1} \), the compositions \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \) should return the original input, creating a perfect balance.

Undoing Process

The undoing process is a method to find the inverse of a function by systematically reversing each step of the original function's operations. It's like following the breadcrumbs back to the starting point in a trail, reversing every decision you made.

Let's take a closer look at how this works using our example function \( f(x) = \frac{2}{5}x + \frac{1}{3} \). To undo it:

• Original operations: First, you multiply \( x \) by \( \frac{2}{5} \), then add \( \frac{1}{3} \).
• Reverse steps: Begin by undoing the addition by subtracting \( \frac{1}{3} \), and then undo the multiplication by multiplying with the reciprocal, \( \frac{5}{2} \).

This process reveals that the inverse function \( f^{-1}(x) \) is \( \frac{5}{2}x - \frac{5}{6} \). It’s a systematic approach, making sure every step in the forward direction has a corresponding backward step.

Verification of Functions

Once you have hypothesized an inverse, the next crucial step is verification. This ensures the correctness of the inverse function you found. Without verification, there's no guarantee your reverse operations work as intended.

To verify an inverse function \( f^{-1} \) of \( f \), you perform two main checks:

• Compute \( f(f^{-1}(x)) \) and ensure it equals \( x \). This checks if applying \( f^{-1} \) and then \( f \) gets you back to where you started.
• Compute \( f^{-1}(f(x)) \) and ensure it equals \( x \). This checks if applying \( f \) and then \( f^{-1} \) also returns you to the original value.

In our example, verifying \( \left(f \circ f^{-1}\right)(x) = x \) used substitution into \( f \):
\[ f(f^{-1}(x)) = f\left( \frac{5}{2}x - \frac{5}{6} \right) = x \]. Similarly, \( \left(f^{-1} \circ f\right)(x) = x \) was checked by substitution into \( f^{-1} \):
\[ f^{-1}(f(x)) = \frac{5}{2}\left( \frac{2}{5}x + \frac{1}{3} - \frac{1}{3} \right) = x \].
These verifications confirm that the inverse function correctly undoes the original, providing a reliable function pair.