Let g(x) be a linear function and f(x)= array|c|c| g(x) & ,x 0 ((1+x)/(2+x))^(1)/(x) & ,x…

Question

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

Accepted Answer

Let $g(x)=a x+b$

Now function $$\mathrm{f}(\mathrm{x})$$ in continuous at $$\mathrm{x}=0$$

$$\begin{aligned}
& \therefore \lim _\limits{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \
& \Rightarrow \lim _\limits{\mathrm{x} \rightarrow 0}\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}}=\mathrm{b} \
& \Rightarrow 0=\mathrm{b} \
& \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax}
\end{aligned}$$

Now, for $$\mathrm{x}>0$$

$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \
& +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \
& \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \
& \text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \
& \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \
& \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \
& =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)
\end{aligned}$$

Solutions · 1