1. Differentiate $g(x)$

Given

Differentiate using the chain rule:

and

So,

2. Use the condition $f''(x)>0$ on $(0,3)$

Since

we know that $f'(x)$ is strictly increasing on $(0,3)$.

Therefore, the sign of

depends on comparing $\dfrac{x}{3}$ and $3-x$.

• If $\dfrac{x}{3}<3-x$, then because $f'$ is increasing,

  $$f'\left(\frac{x}{3}\right)<f'(3-x)$$

  so

  $$g'(x)<0.$$

• If $\dfrac{x}{3}>3-x$, then

  $$f'\left(\frac{x}{3}\right)>f'(3-x)$$

  so

  $$g'(x)>0.$$

The transition point occurs when

3. Find $\alpha$

Solve:

Thus,

Indeed,

• for $x<\frac{9}{4}$, we have $\frac{x}{3}<3-x$, so $g'(x)<0$ and $g$ is decreasing;
• for $x>\frac{9}{4}$, we have $\frac{x}{3}>3-x$, so $g'(x)>0$ and $g$ is increasing.

This matches the given condition.

4. Compute $8\alpha$

5. Check options

The correct option is:

which is Option C.