Questions: For which pair of functions is (f ◦ g)(x)=x? f(x)=x^2 and g(x)=1/x f(x)=2/x and g(x)=2/x f(x)=(x-2)/3 and g(x)=2-3x f(x)=1/2 x-2 and g(x)=1/2 x+2

Transcript text: For which pair of functions is $(f \circ g)(x)=x$ ? $f(x)=x^{2}$ and $g(x)=\frac{1}{x}$ $f(x)=\frac{2}{x}$ and $g(x)=\frac{2}{x}$ $f(x)=\frac{x-2}{3}$ and $g(x)=2-3 x$ $f(x)=\frac{1}{2} x-2$ and. $g(x)=\frac{1}{2} x+2$

Solution

Solution Steps

Step 1: Check Pair $ (f_1, g_1) $

For the functions $ f_1(x) = x^2 $ and $ g_1(x) = \frac{1}{x} $: \[ (f_1 \circ g_1)(x) = f_1(g_1(x)) = f_1\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \] Since $ \frac{1}{x^2} \neq x $, this pair does not satisfy $ (f \circ g)(x) = x $.

Step 2: Check Pair $ (f_2, g_2) $

For the functions $ f_2(x) = \frac{2}{x} $ and $ g_2(x) = \frac{2}{x} $: \[ (f_2 \circ g_2)(x) = f_2(g_2(x)) = f_2\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \] Since $ (f_2 \circ g_2)(x) = x $, this pair satisfies $ (f \circ g)(x) = x $.

Step 3: Check Pair $ (f_3, g_3) $

For the functions $ f_3(x) = \frac{x-2}{3} $ and $ g_3(x) = 2 - 3x $: \[ (f_3 \circ g_3)(x) = f_3(g_3(x)) = f_3(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{-3x}{3} = -x \] Since $ -x \neq x $, this pair does not satisfy $ (f \circ g)(x) = x $.

Final Answer

The only pair of functions that satisfies $ (f \circ g)(x) = x $ is $ (f_2, g_2) $. Thus, the answer is: \[ \boxed{(f(x) = \frac{2}{x}, g(x) = \frac{2}{x})} \]

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