Let g(x)=3 fleft(frac{x}{3}right)+f(3-x) and | vedclass

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(C) Given $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x) > 0$ for all $x \in(0,3)$.Since $f^{\prime \prime}(x) > 0$,$f^{\prime}(x

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$18$

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(C) Given $g(x)=3 f\left(\frac{x}{3}ight)+f(3-x)$ and $f^{\prime \prime}(x) > 0$ for all $x \in(0,3)$.Since $f^{\prime \prime}(x) > 0$,$f^{\prime}(x)$ is an increasing function.Now,differentiate $g(x)$ with respect to $x$:$g^{\prime}(x) = 3 	imes \frac{1}{3} f^{\prime}\left(\frac{x}{3}ight) - f^{\prime}(3-x) = f^{\prime}\left(\frac{x}{3}ight) - f^{\prime}(3-x)$.For $g$ to be decreasing in $(0, \alpha)$,we must have $g^{\prime}(x) $f^{\prime}\left(\frac{x}{3}ight) - f^{\prime}(3-x) Since $f^{\prime}(x)$ is an increasing function,this implies $\frac{x}{3} Solving for $x$: $x + \frac{x}{3} Thus,$\alpha = \frac{9}{4}$.Finally,$8 \alpha = 8 	imes \frac{9}{4} = 18$.

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

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