Answer :
To determine which pair of functions satisfies the condition [tex]\((f \circ g)(x) = x\)[/tex], we need to compute the compositions [tex]\(f(g(x))\)[/tex] for each pair and check if they simplify to [tex]\(x\)[/tex].
### Pair 1: [tex]\(f(x) = x^2\)[/tex] and [tex]\(g(x) = \frac{1}{x}\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex] to hold:
[tex]\[ \frac{1}{x^2} = x \][/tex]
This equation is not true for all [tex]\(x\)[/tex].
### Pair 2: [tex]\(f(x) = \frac{2}{x}\)[/tex] and [tex]\(g(x) = \frac{2}{x}\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = \frac{2 \cdot x}{2} = x \][/tex]
In this case, [tex]\((f \circ g)(x) = x\)[/tex] holds.
### Pair 3: [tex]\(f(x) = \frac{x-2}{3}\)[/tex] and [tex]\(g(x) = 2 - 3x\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{2 - 3x - 2}{3} = \frac{-3x}{3} = -x \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex]:
[tex]\[ -x = x \][/tex]
This is not true for all [tex]\(x\)[/tex].
### Pair 4: [tex]\(f(x) = \frac{1}{2}x - 2\)[/tex] and [tex]\(g(x) = \frac{1}{2}x + 2\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}\left(\frac{1}{2}x + 2\right) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1 \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex]:
[tex]\[ \frac{1}{4}x - 1 = x \][/tex]
This is not true for all [tex]\(x\)[/tex].
Given the above computations, the pair of functions that satisfies [tex]\((f \circ g)(x) = x\)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
Thus, the correct pair is the second pair.
### Pair 1: [tex]\(f(x) = x^2\)[/tex] and [tex]\(g(x) = \frac{1}{x}\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex] to hold:
[tex]\[ \frac{1}{x^2} = x \][/tex]
This equation is not true for all [tex]\(x\)[/tex].
### Pair 2: [tex]\(f(x) = \frac{2}{x}\)[/tex] and [tex]\(g(x) = \frac{2}{x}\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = \frac{2 \cdot x}{2} = x \][/tex]
In this case, [tex]\((f \circ g)(x) = x\)[/tex] holds.
### Pair 3: [tex]\(f(x) = \frac{x-2}{3}\)[/tex] and [tex]\(g(x) = 2 - 3x\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{2 - 3x - 2}{3} = \frac{-3x}{3} = -x \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex]:
[tex]\[ -x = x \][/tex]
This is not true for all [tex]\(x\)[/tex].
### Pair 4: [tex]\(f(x) = \frac{1}{2}x - 2\)[/tex] and [tex]\(g(x) = \frac{1}{2}x + 2\)[/tex]
Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}\left(\frac{1}{2}x + 2\right) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1 \][/tex]
For [tex]\((f \circ g)(x) = x\)[/tex]:
[tex]\[ \frac{1}{4}x - 1 = x \][/tex]
This is not true for all [tex]\(x\)[/tex].
Given the above computations, the pair of functions that satisfies [tex]\((f \circ g)(x) = x\)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
Thus, the correct pair is the second pair.
