Problem 38 - Algebra for College Students

Question

(a) find \(f^{-1}\) and (b) verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). $$ f(x)=2 x-1 $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{x + 1}{2} \), and both composite functions return \( x \).

Step 1: Understand the Function

The function given is \( f(x) = 2x - 1 \). This is a linear function, and our first task is to find its inverse \( f^{-1} \).

Step 2: Replace f(x) with y for Inversion

Set \( y = f(x) = 2x - 1 \). This equation expresses \( y \) in terms of \( x \), and we will use it to find \( x \) in terms of \( y \).

Step 3: Solve for x

Rearrange the equation \( y = 2x - 1 \) to isolate \( x \). Add 1 to both sides: \( y + 1 = 2x \). Then divide both sides by 2: \( x = \frac{y + 1}{2} \).

Step 4: Write the Inverse Function

Rewrite the expression for \( x \) in terms of \( y \) as the inverse function. Thus, \( f^{-1}(x) = \frac{x + 1}{2} \).

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

To verify, substitute \( f^{-1}(x) = \frac{x + 1}{2} \) into \( f(x) \):\[ f\left(f^{-1}(x)\right) = f\left(\frac{x + 1}{2}\right) = 2\left(\frac{x + 1}{2}\right) - 1 \].Simplify: \( x + 1 - 1 = x \). Thus, \( (f \circ f^{-1})(x) = x \).

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

To verify, substitute \( f(x) = 2x - 1 \) into \( f^{-1}(x) \):\[ f^{-1}\left(f(x)\right) = f^{-1}(2x - 1) = \frac{(2x - 1) + 1}{2} \].Simplify: \( \frac{2x - 1 + 1}{2} = \frac{2x}{2} = x \). Hence, \( (f^{-1} \circ f)(x) = x \).

Key Concepts

Linear Functions

Linear functions are among the simplest types of functions that you will encounter in mathematics. They are represented by expressions of the form \( f(x) = ax + b \), where \( a \) and \( b \) are constants. These functions produce straight lines when graphed, with \( a \) being the slope and \( b \) the y-intercept. For example, given the function \( f(x) = 2x - 1 \), the slope \( a \) is 2, indicating the line rises by 2 units for every 1 unit it moves to the right. The y-intercept \( b \) is -1, meaning the line crosses the y-axis at -1.
If you want to find the inverse of a linear function, you basically 'undo' the operations performed by \( f(x) \). Practically, it involves switching the roles of \( x \) and \( y \) in the equation. Then, solve for \( x \) to express it in terms of a new variable, which leads you to the inverse function \( f^{-1}(x) \). For the given function \( f(x) = 2x - 1 \), its inverse is \( f^{-1}(x) = \frac{x + 1}{2} \). This means that if you apply \( f(x) \), you can apply \( f^{-1}(x) \), and get back your original value.

Function Composition

In mathematics, function composition refers to applying one function to the results of another. This is noted as \((f \, \circ \, g)(x)\), which means applying \( f \) to \( g(x) \). When dealing with inverses, two specific compositions are useful for verifying inverse relationships: \( (f \circ f^{-1})(x) \) and \( (f^{-1} \circ f)(x) \).
For instance, if you have the function \( f(x) = 2x - 1 \) and its inverse \( f^{-1}(x) = \frac{x + 1}{2} \), composing these in one order gives you \((f \circ f^{-1})(x) = x\). Composing them in the opposite order \((f^{-1} \circ f)(x)\) will also result in \(x\). This dual composition yielding \(x\) is a signal that two functions are indeed inverses of each other.

• Apply \(f^{-1}(x)\) to an \(x\) value, and you'll get the initial input that you'd need to apply \(f(x)\) to reach the same \(x\).
• Swapping the order works identically: \(f^{-1}\) undoes what \(f\) initially did.

Verification of Inverses

Verification of inverses is an important step to confirm that two functions truly "undo" each other's actions. Simply finding an expression doesn't guarantee it's correct; you need verification. To verify if \( f(x) = 2x - 1 \) and \( f^{-1}(x) = \frac{x + 1}{2} \) are inverses, compute:
1. \( (f \circ f^{-1})(x) = f(f^{-1}(x)) = x \)
2. \( (f^{-1} \circ f)(x) = f^{-1}(f(x)) = x \)
Check both orders, because proper inverses lead precisely back to \(x\). Through substitution, simplification, and getting x back, you ensure these functions genuinely are inverses.

• In the forward direction, plug \( f^{-1}(x) \) into \( f(x) \), simplify, and recover \( x \).
• For the reverse, apply \( f(x) \) inside \( f^{-1}(x) \), and similarly retrieve \( x \).

When these steps result in \(x\), your verification of inverses is complete. This ensures reliable transformations back and forth between the input and output values.