Mathematics - Functions Question with Solution | TestHub

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

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

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

$\Rightarrow f^4\left(x\right)=f\left(\frac{x+1}{1-x}\right)=-\frac{1}{x}$

$\Rightarrow f^6\left(x\right)=-\frac{1}{x}\Rightarrow f^6\left(6\right)=-\frac{1}{6}$

So $f^7\left(x\right)=\left(-\frac{1}{x}\right)=\frac{x+1}{1-x}$

$\Rightarrow f^7\left(7\right)=\frac{8}{-6}=-\frac{4}{3}$

$\therefore \left(-\frac{1}{6}\right)+\left(-\frac{4}{3}\right)=-\frac{3}{2}$