# Functions2

If f(x1) + f(x2) = f{(x1+x2)/(1+x1x2)} , find f(x).

• Manish Shankar ·

Function is log[(1+x)/(1-x)]

• man111 singh ·

\hspace{-20}$Yes Manish Sir...\\\\\\ Then Question defined as follows....\\\\\\ If$\bf{f:\left(-1,1\right)\rightarrow R}$and$\bf{f(x+y) = f\left(\frac{x+y}{1+xy}\right)}$and$\bf{f^{'}(0)=-2}$\\\\\\ Then find value of$\bf{f(x)\;\;,}$where$\bf{x,y\in (-1,1)}$\\\\\\$\bf{\underline{Solution:}}$Given$\bf{f(x+y)=f\left(\frac{x+y}{1+xy}\right)}$and$\bf{f^{'}(0)=-2}$\\\\\\ Diff. both side w.r to$\bf{x\;,}$and$\bf{y}$is treated as a constant.\\\\\\$\bf{f^{'}(x)=f^{'}\left(\frac{x+y}{1+xy}\right)\cdot \frac{d}{dx}\left(\frac{x+y}{1+xy}\right)=f^{'}\left(\frac{x+y}{1+xy}\right)\cdot \left(\frac{1-y^2}{1+xy}\right)^2..........(1)}$\\\\\\ Similarly Diff. both side w.r to$\bf{y\;,}$and$\bf{x}$is treated as a constant.\\\\\\$\bf{f^{'}(y)=f^{'}\left(\frac{x+y}{1+xy}\right)\cdot \frac{d}{dy}\left(\frac{x+y}{1+xy}\right)=f^{'}\left(\frac{x+y}{1+xy}\right)\cdot \left(\frac{1-x^2}{1+xy}\right)^2..........(2)}$\\\\\\ Now Divide$\bf{(1)}$by$\bf{(2)\;,}$we get\\\\\\ \hspace{-20}\bf{\frac{f^{'}(x)}{f^{'}(y)} = \frac{1-y^2}{1-x^2}\Rightarrow (1-x^2)\cdot f^{'}(x)=(1-y^2)\cdot f^{'}(y)}$\\\\\\ So Let $\bf{(1-x^2)\cdot f^{'}(x)=(1-y^2)\cdot f^{'}(y)=k}$\\\\\\ Now Let $\bf{(1-x^2)\cdot f^{'}(x)=k\Rightarrow f^{'}(x)=\frac{k}{1-x^2}}$\\\\\\ Using $\bf{f^{'}(0)=-2\;,}$ So we get $\bf{k=-2}$\\\\\\Now Integrate both side w. r.to $\bf{x\;,}$ we get\\\\\\ $\bf{\Rightarow \int f^{'}(x)=2\int\frac{1}{x^2-1}dx = 2\cdot \frac{1}{2}\left\{\ln \left|1-x|\right-\ln\left|1+x\right|\right\}+\mathbb{C}}$\\\\\\ So $\bf{f(x)= \ln \left|\frac{1-x}{1+x}\right|+\mathbb{C}}$\\\\\\ Now Put $\bf{x=y=0}$ in original equation, we get $\bf{f(0)=0}$\\\\\\ So we get $\bf{C=0}$\\\\\\ So $\bf{f(x)=\ln \left|\frac{1-x}{1+x}\right|\;\;,}$ where $\bf{x,y\in (-1,1)}$

Sushovan Halder ·

manish sir, how to solve this?

• man111 singh ·

\hspace{-16}$May be He/She is asking for.........\\\\\\ If$\bf{f(x_{1}+x_{2}) = f\left(\frac{x_{1}+x_{2}}{1-x_{1}\cdot x_{2}}\right)\;,}$Then$\bf{f(x)}\$ is