Question
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Find \( f(g(x)) \) and \( g(f(x)) \) and determine whether the pair of functions \( f \) and \( g \) are inverses of each other \( f(x)=4 x+2 \) and \( g(x)=\frac{x-2}{4} \) a. \( f(g(x))=x \) (Simplify your answer) b. \( g(f(x))=\square \) (Simplify your answer)

Ask by Garrett Hall. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\( f(g(x)) = x \) and \( g(f(x)) = x \). Therefore, \( f \) and \( g \) are inverses of each other.

Solution

Let's determine \( f(g(x)) \) and \( g(f(x)) \) for the given functions and verify whether \( f \) and \( g \) are inverses of each other. **Given Functions:** \[ f(x) = 4x + 2 \\ g(x) = \frac{x - 2}{4} \] ### a. Compute \( f(g(x)) \) \[ f(g(x)) = f\left(\frac{x - 2}{4}\right) \] \[ = 4 \left( \frac{x - 2}{4} \right) + 2 \] \[ = (x - 2) + 2 \] \[ = x \] So, \[ f(g(x)) = x \] ### b. Compute \( g(f(x)) \) \[ g(f(x)) = g(4x + 2) \] \[ = \frac{(4x + 2) - 2}{4} \] \[ = \frac{4x}{4} \] \[ = x \] So, \[ g(f(x)) = x \] ### Conclusion: Are \( f \) and \( g \) Inverses? Since both compositions satisfy: \[ f(g(x)) = x \quad \text{and} \quad g(f(x)) = x \] this means that \( f \) and \( g \) are indeed **inverse functions** of each other. **Answer Summary:** - **a.** \( f(g(x)) = x \) - **b.** \( g(f(x)) = x \) Since both compositions equal \( x \), \( f \) and \( g \) are inverses of one another.

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The Deep Dive

To find \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \): 1. First, we have \( g(x) = \frac{x-2}{4} \). 2. Now substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{x-2}{4}\right) = 4\left(\frac{x-2}{4}\right) + 2 \] Simplifying this: \[ = (x - 2) + 2 = x \] So, \( f(g(x)) = x \). Now let's find \( g(f(x)) \): 1. We start with \( f(x) = 4x + 2 \). 2. Substituting \( f(x) \) into \( g(x) \): \[ g(f(x)) = g(4x + 2) = \frac{(4x + 2) - 2}{4} \] Simplifying this: \[ = \frac{4x}{4} = x \] So, \( g(f(x)) = x \). Since \( f(g(x)) = x \) and \( g(f(x)) = x \), the functions \( f \) and \( g \) are indeed inverses of each other!

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