The question asks for which pair of functions $(f \circ g)(x) = x$, which means the composition of functions $f$ and $g$ results in $x$.

We need to check each pair by calculating $(f \circ g)(x)$, which means applying $g(x)$ first, then applying $f$ to the result of $g(x)$.

Let’s evaluate the options one by one:

Option A:

• $f(x) = x^2$ and $g(x) = \frac{1}{x}$
• $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}$
• This is not equal to $x$.

Option B:

• $f(x) = \frac{2}{x}$ and $g(x) = \frac{2}{x}$
• $(f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x$
• This satisfies $(f \circ g)(x) = x$.

Option C:

• $f(x) = \frac{x - 2}{3}$ and $g(x) = 2 - 3x$
• $(f \circ g)(x) = f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{-3x}{3} = -x$
• This is not equal to $x$.

Option D:

• $f(x) = \frac{1}{2}x - 2$ and $g(x) = \frac{1}{2}x + 2$
• $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}\left(\frac{1}{2}x + 2\right) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1$
• This is not equal to $x$.

Conclusion:

The correct pair of functions is Option B: $f(x) = \frac{2}{x} \text{ and } g(x) = \frac{2}{x}.$

Would you like more details on any of these steps or have further questions?

Here are five related questions for further exploration:

1. How do we define function composition?
2. What is the domain of a composite function?
3. How do we find the inverse of a function?
4. What are some real-life applications of composite functions?
5. How do you verify if two functions are inverses?

Tip: When working with composite functions, always apply the inner function first and then the outer function.