Problem 26 - Intermediate Algebra

Question

Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=-\frac{3}{4} x+\frac{1}{3}\) and \(g(x)=-\frac{4}{3} x+\frac{4}{9}\)

Step-by-Step Solution

Answer

The functions are inverses: \( (f \circ g)(x) = x \) and \( (g \circ f)(x) = x \).

Step 1: Substitute and Simplify for \( (f \circ g)(x) \)

Start by substituting \( g(x) = -\frac{4}{3}x + \frac{4}{9} \) into \( f(x) \). Compute \( f(g(x)) = f\left(-\frac{4}{3}x + \frac{4}{9}\right) \). This gives:\[ f\left(-\frac{4}{3}x + \frac{4}{9}\right) = -\frac{3}{4}\left(-\frac{4}{3}x + \frac{4}{9}\right) + \frac{1}{3} \]Simplify inside the function first:\[ -\frac{3}{4}\left(-\frac{4}{3}x + \frac{4}{9}\right) = x - \frac{1}{3} \]Adding \( \frac{1}{3} \):\[ f\left(-\frac{4}{3}x + \frac{4}{9}\right) = x - \frac{1}{3} + \frac{1}{3} = x \]Thus, \( (f \circ g)(x) = x \).

Step 2: Substitute and Simplify for \( (g \circ f)(x) \)

Now substitute \( f(x) = -\frac{3}{4}x + \frac{1}{3} \) into \( g(x) \). Compute \( g(f(x)) = g\left(-\frac{3}{4}x + \frac{1}{3}\right) \). This gives:\[ g\left(-\frac{3}{4}x + \frac{1}{3}\right) = -\frac{4}{3}\left(-\frac{3}{4}x + \frac{1}{3}\right) + \frac{4}{9} \]Simplify inside the function first:\[ -\frac{4}{3}\left(-\frac{3}{4}x + \frac{1}{3}\right) = x - \frac{4}{9} \]Adding \( \frac{4}{9} \):\[ g\left(-\frac{3}{4}x + \frac{1}{3}\right) = x - \frac{4}{9} + \frac{4}{9} = x \]Thus, \( (g \circ f)(x) = x \).

Step 3: Conclusion of Inverse Function Verification

Since both substitutes resulted in simplifying to \( x \), we have shown: \( (f \circ g)(x) = x \) and \( (g \circ f)(x) = x \). Therefore, the functions \( f(x) = -\frac{3}{4}x + \frac{1}{3} \) and \( g(x) = -\frac{4}{3}x + \frac{4}{9} \) are inverses of each other.

Key Concepts

Function Composition

Understanding function composition is crucial when working with inverse functions. It involves combining two functions to form a single new function.
Defined as \((f \circ g)(x)\),
it means applying function \(g\) to \(x\) first, then applying function \(f\) to the result of \(g(x)\).

• Start by writing the expression for \((f \circ g)(x) = f(g(x))\).
• Replace \(g(x)\) with its expression.
• Simplify the resulting equation step by step until reduced.

In the given exercise, we check if \((f \circ g)(x) = x\)
to ensure the functions are inverse. Observe how substitution and simplification play out.
This method allows us to verify if the functions work together to unravel each other back to \(x\).

Linear Functions

Linear functions are foundational in mathematics. They map inputs to outputs in a straight-line manner. Expressed as \(f(x) = mx + b\), where *m* is the slope and *b* is the y-intercept. Investigating the linear nature:

• Slope (*m*): Indicates how much \(y\) changes for each unit change in \(x\).
• Y-intercept (*b*): The point where the line crosses the y-axis when \(x = 0\).

A simple form makes calculations in function composition easier to manage. Both functions, \(f(x) = -\frac{3}{4}x + \frac{1}{3}\) and \(g(x) = -\frac{4}{3}x + \frac{4}{9}\), have slopes and y-intercepts that play a role in reaching an identity function. Their linear nature allows for straightforward substitution and arithmetic manipulation.

Inverse Verification

Verifying inverse functions is the next important step, ensuring two functions truly undo each other. This entails showing that when one function is applied after the other, the result is just \(x\).

• Perform function compositions in both orders: \((f \circ g)(x)\) and \((g \circ f)(x)\).
• Substitute the expressions properly, one inside the other.
• Simplify to check if the result is \(x\).

In our exercise, we saw that combining \(f(x)\) into \(g(x)\) and \(g(x)\) into \(f(x)\) both yield \(x\):
Thus, confirming that they are inverses.
This symmetry implies a perfect cancelation where each function offsets the other’s transformation of any input \(x\). So, these functions reverse effects.