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Sagot :
To determine for which pair of functions [tex]\((f \circ g)(x) = x\)[/tex], let's analyze each given pair of functions step by step.
### Pair 1: [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{1}{x} \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{x^2} \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex].
### Pair 2: [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{2}{x} \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = \frac{2x}{2} = x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) = x \)[/tex] for the pair [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex].
### Pair 3: [tex]\( f(x) = \frac{x - 2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f(2 - 3x) \][/tex]
2. Substitute [tex]\( g(x) = 2 - 3x \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{2 - 3x - 2}{3} = \frac{-3x}{3} = -x \][/tex]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ -x \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = \frac{x - 2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex].
### Pair 4: [tex]\( f(x) = \frac{1}{2}x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2}x + 2 \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2} x + 2\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{1}{2} x + 2 \)[/tex] into [tex]\( f \)[/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]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{4} x - 1 \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = \frac{1}{2} x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2} x + 2 \)[/tex].
### Summary
The only pair of functions for which [tex]\( (f \circ g)(x) = x \)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
### Pair 1: [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{1}{x} \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \][/tex]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{x^2} \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \frac{1}{x} \)[/tex].
### Pair 2: [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{2}{x} \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = \frac{2x}{2} = x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) = x \)[/tex] for the pair [tex]\( f(x) = \frac{2}{x} \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex].
### Pair 3: [tex]\( f(x) = \frac{x - 2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f(2 - 3x) \][/tex]
2. Substitute [tex]\( g(x) = 2 - 3x \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{2 - 3x - 2}{3} = \frac{-3x}{3} = -x \][/tex]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ -x \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = \frac{x - 2}{3} \)[/tex] and [tex]\( g(x) = 2 - 3x \)[/tex].
### Pair 4: [tex]\( f(x) = \frac{1}{2}x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2}x + 2 \)[/tex]
1. Composition [tex]\( f \circ g \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2} x + 2\right) \][/tex]
2. Substitute [tex]\( g(x) = \frac{1}{2} x + 2 \)[/tex] into [tex]\( f \)[/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]
3. Compare with [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{4} x - 1 \neq x \][/tex]
Conclusion: [tex]\( (f \circ g)(x) \neq x \)[/tex] for the pair [tex]\( f(x) = \frac{1}{2} x - 2 \)[/tex] and [tex]\( g(x) = \frac{1}{2} x + 2 \)[/tex].
### Summary
The only pair of functions for which [tex]\( (f \circ g)(x) = x \)[/tex] is:
[tex]\[ f(x) = \frac{2}{x} \quad \text{and} \quad g(x) = \frac{2}{x} \][/tex]
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