Let \( f(x)=\frac{1}{x-2} \) and \( g(x)=\frac{6}{x}+2 \) Find the following functions. Simplify your answers. \( f(g(x))=\square \) \( g(f(x))=\square \)

Evaluate the expression by following steps: - step0: Evaluate: \(\frac{6}{x}+2\) - step1: Substitute: \(\frac{6}{\frac{1}{x-2}}+2\) - step2: Multiply by the reciprocal: \(6\left(x-2\right)+2\) - step3: Expand the expression: \(6x-12+2\) - step4: Add the numbers: \(6x-10\) Substitute \( x=\frac{6}{x}+2 \) into the expression \( \frac{1}{x-2} \). Evaluate the expression by following steps: - step0: Evaluate: \(\frac{1}{x-2}\) - step1: Substitute: \(\frac{1}{\frac{6}{x}+2-2}\) - step2: Add the terms: \(\frac{1}{\frac{6+2x}{x}-2}\) - step3: Subtract the terms: \(\frac{1}{\frac{6}{x}}\) - step4: Multiply by the reciprocal: \(1\times \frac{x}{6}\) - step5: Multiply: \(\frac{x}{6}\) Let's find the compositions of the functions step by step. 1. **Finding \( f(g(x)) \)**: - We substitute \( g(x) = \frac{6}{x} + 2 \) into \( f(x) = \frac{1}{x-2} \): \[ f(g(x)) = f\left(\frac{6}{x} + 2\right) = \frac{1}{\left(\frac{6}{x} + 2\right) - 2} \] - Simplifying the expression: \[ f(g(x)) = \frac{1}{\frac{6}{x} + 2 - 2} = \frac{1}{\frac{6}{x}} = \frac{x}{6} \] 2. **Finding \( g(f(x)) \)**: - We substitute \( f(x) = \frac{1}{x-2} \) into \( g(x) = \frac{6}{x} + 2 \): \[ g(f(x)) = g\left(\frac{1}{x-2}\right) = \frac{6}{\frac{1}{x-2}} + 2 \] - Simplifying the expression: \[ g(f(x)) = 6(x-2) + 2 = 6x - 12 + 2 = 6x - 10 \] Thus, the simplified results are: - \( f(g(x)) = \frac{x}{6} \) - \( g(f(x)) = 6x - 10 \)