Certainly! Let's analyze each pair of functions to determine whether their composition [tex]\((f \circ g)(x) = x\)[/tex].
### 1. [tex]\(f(x) = x^2\)[/tex] and [tex]\(g(x) = \frac{1}{x}\)[/tex]
We need to compute [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)
\][/tex]
Substitute [tex]\(g(x) = \frac{1}{x}\)[/tex]:
[tex]\[
f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}
\][/tex]
This does not equal [tex]\(x\)[/tex], so [tex]\((f \circ g)(x) \ne x\)[/tex].
### 2. [tex]\(f(x) = \frac{2}{x}\)[/tex] and [tex]\(g(x) = \frac{2}{x}\)[/tex]
We need to compute [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right)
\][/tex]
Substitute [tex]\(g(x) = \frac{2}{x}\)[/tex]:
[tex]\[
f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x
\][/tex]
This equals [tex]\(x\)[/tex], so [tex]\((f \circ g)(x) = x\)[/tex].
### 3. [tex]\(f(x) = \frac{x-2}{3}\)[/tex] and [tex]\(g(x) = 2 - 3x\)[/tex]
We need to compute [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[
(f \circ g)(x) = f(g(x)) = f(2 - 3x)
\][/tex]
Substitute [tex]\(g(x) = 2 - 3x\)[/tex]:
[tex]\[
f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{2 - 3x - 2}{3} = \frac{-3x}{3} = -x
\][/tex]
This does not equal [tex]\(x\)[/tex], so [tex]\((f \circ g)(x) \ne x\)[/tex].
### 4. [tex]\(f(x) = \frac{1}{2}x - 2\)[/tex] and [tex]\(g(x) = \frac{1}{2}x + 2\)[/tex]
We need to compute [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[
(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2}x + 2\right)
\][/tex]
Substitute [tex]\(g(x) = \frac{1}{2}x + 2\)[/tex]:
[tex]\[
f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}\left(\frac{1}{2}x + 2\right) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1
\][/tex]
This does not equal [tex]\(x\)[/tex], so [tex]\((f \circ g)(x) \ne x\)[/tex].
### Conclusion
The pair of functions for which [tex]\((f \circ g)(x) = x\)[/tex] is:
[tex]\[
f(x)=\frac{2}{x} \text{ and } g(x)=\frac{2}{x}
\][/tex]
