Let f(x)=frac{x-1}{x+1}, x in R-{0,-1,1). If | vedclass

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Question

$f(x)=\frac{x-1}{x+1}$

$\Rightarrow f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$

$f^{3}(x)=f\left(f^{2}(x)\right)=

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

$f(x)=\frac{x-1}{x+1}$

$\Rightarrow f^{2}(x)=f(f(x))=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$

$f^{3}(x)=f\left(f^{2}(x)ight)=f\left(-\frac{1}{x}ight)=\frac{x+1}{1-x}$

$\Rightarrow f^{4}(x)=f\left(\frac{x+1}{1-x}ight)=-\frac{1}{x}$

$\Rightarrow f^{6}(x)=-\frac{1}{x} \Rightarrow f^{6}(6)=-\frac{1}{8}$

$f^{7}(x)=\left(-\frac{1}{x}ight)=\frac{x+1}{1-x}$

$\Rightarrow f^{7}(7)=\frac{8}{-6}=-\frac{4}{3}$

$	herefore-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}$

Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{a+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{\prime}(6)+f(7)$ is equal to

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