For which pair of functions is $$ (f \circ g)(x)=x $$ ? $$ f(x)=x^{2} \text { and } g(x)=\frac{1}{x} $$ $$ f(x)=\frac{2}{x} $$ and $$ g(x)=\frac{2}{x} $$ $$ f(x)=\frac{x-2}{3} $$ and $$ g(x)=2-3 x $$ $$ f(x)=\frac{1}{2} x-2 $$ and $$ g(x)=\frac{1}{2} x+2 $$

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To determine which pair of functions satisfies the condition $$ (f \circ g)(x) = x $$, we need to evaluate $$ f(g(x)) $$ for each pair of functions and check if it equals $$ x $$.

1. **For $$ f(x) = x^{2} $$ and $$ g(x) = \frac{1}{x} $$**:
   $$
   f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^{2} = \frac{1}{x^{2}}
   $$

2. **For $$ f(x) = \frac{2}{x} $$ and $$ g(x) = \frac{2}{x} $$**:
   $$
   f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x
   $$

3. **For $$ f(x) = \frac{x-2}{3} $$ and $$ g(x) = 2-3x $$**:
   $$
   f(g(x)) = f(2-3x) = \frac{(2-3x)-2}{3} = \frac{-3x}{3} = -x
   $$

4. **For $$ f(x) = \frac{1}{2} x - 2 $$ and $$ g(x) = \frac{1}{2} x + 2 $$**:
   $$
   f(g(x)) = f\left(\frac{1}{2} x + 2\right) = \frac{1}{2}\left(\frac{1}{2} x + 2\right) - 2 = \frac{1}{4} x + 1 - 2 = \frac{1}{4} x - 1
   $$

Now, let's summarize the results:

1. $$ f(g(x)) = \frac{1}{x^{2}} $$ (not equal to $$ x $$)
2. $$ f(g(x)) = x $$ (this pair works)
3. $$ f(g(x)) = -x $$ (not equal to $$ x $$)
4. $$ f(g(x)) = \frac{1}{4} x - 1 $$ (not equal to $$ x $$)

The only pair of functions for which $$ (f \circ g)(x) = x $$ is:
- $$ f(x) = \frac{2}{x} $$ and $$ g(x) = \frac{2}{x} $$
The pair $$ f(x) = \frac{2}{x} $$ and $$ g(x) = \frac{2}{x} $$ satisfies $$ (f \circ g)(x) = x $$.

For which pair of functions is ?

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