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We have that R_{g} = (-\infty, \infty) \mathrm{and} D_{f} = \{ \mathbb{R} | x \ne 1 \},
Since R_{g} \nsubseteq D_{f}, fg does not exist.

gf(x) = g( \frac{2+x}{1-x})

= 1 - 2(\frac{2+x}{1-x})

= \frac{1-x-4-2x}{1-x}

= \frac{3x+3}{x-1}

Let (gf)^{-1}(5) = a

gf[(gf)^{-1}(5)] = gf(a)

5 = \frac{3+3a}{a-1}

5a - 5 = 3 + 3a

a = 4

KS Comments:

A very standard question. The last part requires student to understand the relationship between a function and its inverse, instead of simply stating that they are a reflection about y=x.

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