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

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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:
- $$ f(g(x)) = \frac{x}{6} $$
- $$ g(f(x)) = 6x - 10 $$
$$ f(g(x)) = \frac{x}{6} $$ $$ g(f(x)) = 6x - 10 $$

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

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