For x,, in ,R,,x, ne ,0, let {f_0}(x) = | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac {5}{3}$

Vedclass · Answer

${f_1}\left( x ight) = {f_{0 + 1}}\left( x ight) = {f_0}\left( {{f_0}\left( x ight)} ight) = \frac{1}{{1 - \frac{1}{{1 - x}}}} = \frac{{x - 1}}{x}$

${f_2}\left( x ight) = {f_{1 + 1}}\left( x ight) = {f_0}\left( {{f_1}\left( x ight)} ight) = \frac{1}{{1 - \frac{{x - 1}}{x}}} = x$

${f_3}\left( x ight) = {f_{2 + 1}}\left( x ight) = {f_0}\left( {{f_2}\left( x ight)} ight) = {f_0}\left( x ight) = \frac{1}{{1 - x}}$

${f_4}\left( x ight) = {f_{3 + 1}}\left( x ight) = {f_0}\left( {{f_3}\left( x ight)} ight) = \frac{{x - 1}}{x}$

${f_0} = {f_3} = {f_6} = ........ = \frac{1}{{1 - x}}$

${f_1} = {f_4} = {f_7} = ........ = \frac{{x - 1}}{x}$

${f_2} = {f_5} = {f_8} = ........ = x$

${f_{100}}\left( 3 ight) = \frac{{3 - 1}}{3} = \frac{2}{3}{f_1}\left( {\frac{2}{3}} ight) = \frac{{\frac{2}{3} - 1}}{{\frac{2}{3}}} =  - \frac{1}{2}$

            ${f_2}\left( {\frac{3}{2}} ight) = \frac{3}{2}$

$	herefore {f_{100}}\left( 3 ight) + {f_1}\left( {\frac{2}{3}} ight) + {f_2}\left( {\frac{3}{2}} ight) = \frac{5}{3}$

For $x\,\, \in \,R\,,x\, \ne \,0,$ 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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