Let $g\left(x\right)$ be a linear function and $f\left(x\right)=\left\{\begin{array}{cc}g\left(x\right),& x\le 0\\ {\left(\frac{1+x}{2+x}\right)}^{\frac{1}{x}},& x>0\end{array}\right.$, is continuous at $x=0$. If $f^{\text{'}}\left(1\right)=f\left(-1\right)$, then the value of $g\left(3\right)$ is

Let, $g\left(x\right)=ax+b$

Now function $f\left(x\right)$ in continuous at $x=0$

$\Rightarrow \underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{\lim }{\left(\frac{1+x}{2+x}\right)}^{\frac{1}{x}}=b$

$\Rightarrow 0=b$ 

$\Rightarrow g\left(x\right)=ax$

Now, for $x>0$

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

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

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

$\Rightarrow \frac{1}{f\left(x\right)}\times f^{\text{'}}\left(x\right)=\frac{x\left({\displaystyle \frac{1}{1+x}-\frac{1}{2+x}}\right)-\left[\log \left(1+x\right)-\log \left(2+x\right)\right]}{x^2}$

$\Rightarrow f^{\text{'}}\left(x\right)=f\left(x\right)\left\{\frac{x\left({\displaystyle \frac{1}{1+x}-\frac{1}{2+x}}\right)-\left[\log \left(1+x\right)-\log \left(2+x\right)\right]}{x^2}\right\}$

$\Rightarrow f^{\text{'}}\left(1\right)=f\left(1\right)\left\{\frac{\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)-\left[\log \left(2\right)-\log \left(3\right)\right]}{1^2}\right\}$

$\Rightarrow f^{\text{'}}\left(1\right)=\frac{2}{3}\left\{\frac{1}{6}-\log \left(\frac{2}{3}\right)\right\}$

And, $f\left(-1\right)=g\left(-1\right)=-a$

$\Rightarrow a=\frac{2}{3}\log \left(\frac{2}{3}\right)-\frac{1}{9}$

Now,  finding $g\left(3\right)=3a+0$

$\Rightarrow g\left(3\right)=2\log \left(\frac{2}{3}\right)-\frac{1}{3}$

$\Rightarrow g\left(3\right)=\log \left(\frac{4}{9\cdot e^{{\displaystyle \frac{1}{3}}}}\right)$