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

<p>Let $g(x)=a x+b$</p> <p>Now function $\mathrm{f}(\mathrm{x})$ in continuous at $\mathrm{x}=0$</p> <p>$$\begin{aligned} & \therefore \lim _\limits{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \\ & \Rightarrow \lim _\limits{\mathrm{x} \rightarrow 0}\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}}=\mathrm{b} \\ & \Rightarrow 0=\mathrm{b} \\ & \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax} \end{aligned}$$</p> <p>Now, for $\mathrm{x}>0$</p> <p>$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \\ & +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \\ & \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \\ & \text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \\ & \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \\ & \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \\ & =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right) \end{aligned}$$</p>