Let \(g(x)\) be a linear function and \(f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\righ…

Let \(g(x)=ax+b\).

Now function \(f(x)\) is continuous at \(x=0\).

\(\therefore \lim_{x\to 0^+}f(x)=f(0)\)

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

\(\Rightarrow b=0\)

\(\therefore g(x)=ax\)

Now, for \(x>0\),

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

Taking log on both sides,

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

Differentiating both sides,

\(\frac{f'(x)}{f(x)}=\frac{1}{x}\cdot\frac{1}{(1+x)(2+x)}-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)\)

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

\(\therefore f'(1)=\frac{2}{3}\left[\frac{1}{6}-\ln\left(\frac{2}{3}\right)\right]\)

\(\Rightarrow f'(1)=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)

And \(f(-1)=g(-1)=-a\).

\(\therefore -a=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)

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

\(\therefore g(3)=3a\)

\(=2\ln\left(\frac{2}{3}\right)-\frac{1}{3}\)

\(=\ln\left(\frac{4}{9}\right)-\ln e^{\frac{1}{3}}\)

\(=\ln\left(\frac{4}{9e^{\frac{1}{3}}}\right)\)