Let $$g(x) = 3f\left(\frac{x}{3}\right) + f(3 - x)$$ 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

We are told that $$f''(x) \gt 0$$ for every $$x \in (0,3)$$.
A positive second derivative means $$f(x)$$ is strictly convex and its first derivative $$f'(x)$$ is strictly increasing on $$(0,3)$$.

Define $$g(x)=3\,f\!\left(\tfrac{x}{3}\right)+f(3-x)\,.$$

Step 1 : Differentiate $$g(x)$$
Use the chain rule term-by-term:

$$\frac{d}{dx}\Bigl[3\,f\!\left(\tfrac{x}{3}\right)\Bigr] = 3\;f'\!\left(\tfrac{x}{3}\right)\cdot\frac{1}{3} = f'\!\left(\tfrac{x}{3}\right)$$

$$\frac{d}{dx}\Bigl[f(3-x)\Bigr] = f'(3-x)\cdot(-1) = -\,f'(3-x)$$

Therefore

$$g'(x)=f'\!\left(\tfrac{x}{3}\right)-f'(3-x)\,.$$

Step 2 : Locate the point $$\alpha$$ where $$g(x)$$ switches from decreasing to increasing
The description “$$g$$ is decreasing on $$(0,\alpha)$$ and increasing on $$(\alpha,3)$$” means $$g'(x)\lt 0$$ for $$0\lt x\lt\alpha$$, $$g'(\alpha)=0$$, and $$g'(x)\gt 0$$ for $$\alpha\lt x\lt3$$.
Hence $$g'(\alpha)=0$$.

Set $$g'(\alpha)=0$$:

$$f'\!\left(\tfrac{\alpha}{3}\right)-f'(3-\alpha)=0 \;\Longrightarrow\; f'\!\left(\tfrac{\alpha}{3}\right)=f'(3-\alpha)\,.$$

Step 3 : Use the strict monotonicity of $$f'(x)$$
Because $$f'(x)$$ is strictly increasing on $$(0,3)$$, it is a one-to-one (injective) function there.
If two inputs give the same output, the inputs themselves must be equal:

$$\tfrac{\alpha}{3}=3-\alpha\,.$$

Step 4 : Solve for $$\alpha$$

Multiply both sides by $$3$$: $$\alpha = 9-3\alpha$$.
Bring like terms together: $$4\alpha = 9$$.
Hence $$\alpha = \dfrac{9}{4}=2.25$$, which indeed lies inside $$(0,3)$$.

Step 5 : Compute $$8\alpha$$

$$8\alpha = 8 \times \dfrac{9}{4} = 18.$$

Therefore $$8\alpha = 18$$, which corresponds to Option C.