Let g(x)=3f((x)/(3))+f(3-x) and f^''(x)0 for all x (0,3).

Question

Let $g(x)=3f(\frac{x}{3})+f(3-x)$ and ${f}^{''}(x)>0$ for all $x\in (0,3)$. If $g$ is decreasing in $(0,\alpha )$ and increasing in $(\alpha ,3)$, then $8\alpha $ is :

Accepted Answer

$g(x)=3f(\frac{x}{3})+f(3-x){f}^{''}(x)>0\forallx\in (0,3)$

$\Rightarrow {f}^{'}(x)$ is increasing function

$\begin{gathered} {g}^{'}(x)=3\times \frac{1}{3}\cdot {f}^{'}(\frac{x}{3})-{f}^{'}(3-x) \ ={f}^{'}(\frac{x}{3})-{f}^{'}(3-x) \end{gathered}$

If $g$ is decreasing in $(0,\alpha )$

$ $$\& {g}^{'}(x)Therefore $\alpha =\frac{9}{4}$$$

Then $

Solutions · 1