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Let's analyze each pair of functions to determine for which pair the composition [tex]\((f \circ g)(x) = x\)[/tex] and [tex]\((g \circ f)(x) = x\)[/tex] holds true.
### Pair 1: [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g(x^2) = \frac{1}{x^2} \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Pair 2: [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \][/tex]
For this pair, [tex]\((f \circ g)(x) = x\)[/tex] and [tex]\((g \circ f)(x) = x\)[/tex]. So, this pair satisfies the condition.
### Pair 3: [tex]\( f(x) = \frac{x-2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{-3x}{3} = -x \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{x-2}{3}\right) = 2 - 3 \left(\frac{x-2}{3}\right) = 2 - (x - 2) = 4 - x \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Pair 4: [tex]\( f(x) = \frac{1}{2}x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2}x + 2 \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/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]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{1}{2}x - 2\right) = \frac{1}{2}\left(\frac{1}{2}x - 2\right) + 2 = \frac{1}{4}x - 1 + 2 = \frac{1}{4}x + 1 \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Conclusion
The only pair of functions for which [tex]\((f \circ g)(x) = x\)[/tex] and [tex]\((g \circ f)(x) = x\)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
So, the correct answer is the second pair.
### Pair 1: [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g(x^2) = \frac{1}{x^2} \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Pair 2: [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \][/tex]
For this pair, [tex]\((f \circ g)(x) = x\)[/tex] and [tex]\((g \circ f)(x) = x\)[/tex]. So, this pair satisfies the condition.
### Pair 3: [tex]\( f(x) = \frac{x-2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[ f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{-3x}{3} = -x \][/tex]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{x-2}{3}\right) = 2 - 3 \left(\frac{x-2}{3}\right) = 2 - (x - 2) = 4 - x \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Pair 4: [tex]\( f(x) = \frac{1}{2}x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2}x + 2 \)[/tex]
1. Compute [tex]\( (f \circ g)(x) \)[/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]
2. Compute [tex]\( (g \circ f)(x) \)[/tex]:
[tex]\[ g(f(x)) = g\left(\frac{1}{2}x - 2\right) = \frac{1}{2}\left(\frac{1}{2}x - 2\right) + 2 = \frac{1}{4}x - 1 + 2 = \frac{1}{4}x + 1 \][/tex]
For these functions, [tex]\((f \circ g)(x) \neq x\)[/tex] and [tex]\((g \circ f)(x) \neq x\)[/tex].
### Conclusion
The only pair of functions for which [tex]\((f \circ g)(x) = x\)[/tex] and [tex]\((g \circ f)(x) = x\)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
So, the correct answer is the second pair.
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