For x in R, x neq 0, let f_0(x) = frac{1}{1 - | vedclass

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(C) $f_1(x) = f_{0+1}(x) = f_0(f_0(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}$$f_2(x) = f_{1+1}(x) = f_0(f_1(x)) = \frac{1}{1 - \frac{x - 1

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$\frac{5}{3}$

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(C) $f_1(x) = f_{0+1}(x) = f_0(f_0(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}$$f_2(x) = f_{1+1}(x) = f_0(f_1(x)) = \frac{1}{1 - \frac{x - 1}{x}} = x$$f_3(x) = f_{2+1}(x) = f_0(f_2(x)) = f_0(x) = \frac{1}{1 - x}$Since $f_3(x) = f_0(x)$,the function repeats with a period of $3$.$f_{100}(3) = f_{3 	imes 33 + 1}(3) = f_1(3) = \frac{3 - 1}{3} = \frac{2}{3}$$f_1\left( \frac{2}{3} ight) = \frac{\frac{2}{3} - 1}{\frac{2}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{2}$$f_2\left( \frac{3}{2} ight) = \frac{3}{2}$Therefore,$f_{100}(3) + f_1\left( \frac{2}{3} ight) + f_2\left( \frac{3}{2} ight) = \frac{2}{3} - \frac{1}{2} + \frac{3}{2} = \frac{2}{3} + 1 = \frac{5}{3}$

For $x \in R, x \neq 0,$ let $f_0(x) = \frac{1}{1 - x}$ and $f_{n + 1}(x) = f_0(f_n(x)),$ $n = 0, 1, 2, ....$ Then the value of $f_{100}(3) + f_1\left( \frac{2}{3} \right) + f_2\left( \frac{3}{2} \right)$ is equal to

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