Key Concepts and Formulas

• Monotonicity and Derivatives: A function $h(x)$ is decreasing where $h'(x) < 0$ and increasing where $h'(x) > 0$. At a local extremum (like $\alpha$), $h'(\alpha) = 0$.
• Chain Rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$.
• Increasing Derivative and Concavity: If $f''(x) > 0$, then $f'(x)$ is an increasing function.

Step-by-Step Solution

Step 1: Find the derivative of g(x)

We are given $g(x) = 3 f\left(\frac{x}{3}\right) + f(3-x)$. We need to find $g'(x)$ using the chain rule.

$g'(x) = \frac{d}{dx} \left[3 f\left(\frac{x}{3}\right) + f(3-x)\right]$ $g'(x) = 3 \cdot f'\left(\frac{x}{3}\right) \cdot \frac{d}{dx}\left(\frac{x}{3}\right) + f'(3-x) \cdot \frac{d}{dx}(3-x)$ $g'(x) = 3 \cdot f'\left(\frac{x}{3}\right) \cdot \frac{1}{3} + f'(3-x) \cdot (-1)$ $g'(x) = f'\left(\frac{x}{3}\right) - f'(3-x)$

Step 2: Apply the critical point condition

We know that $g(x)$ changes from decreasing to increasing at $x=\alpha$. Therefore, $g'(\alpha) = 0$.

$g'(\alpha) = f'\left(\frac{\alpha}{3}\right) - f'(3-\alpha) = 0$ $f'\left(\frac{\alpha}{3}\right) = f'(3-\alpha)$

Step 3: Use the increasing nature of f'(x)

Since $f''(x) > 0$ for $x \in (0,3)$, $f'(x)$ is an increasing function. Therefore, if $f'(a) = f'(b)$, then $a = b$.

$\frac{\alpha}{3} = 3 - \alpha$

Step 4: Solve for alpha

Solve the equation for $\alpha$:

$\alpha = 9 - 3\alpha$ $4\alpha = 9$ $\alpha = \frac{9}{4}$

Step 5: Calculate 8*alpha

We need to find the value of $8\alpha$.

$8\alpha = 8 \cdot \frac{9}{4} = 2 \cdot 9 = 18$

Common Mistakes & Tips

• Chain Rule Errors: Be careful when applying the chain rule, especially with the negative sign in $f(3-x)$.
• Assuming f'(x) is constant: The key is to recognize that $f'(x)$ is increasing (not necessarily linear or constant), which allows us to equate the arguments.
• Interval Considerations: Always check if the found value of $\alpha$ lies within the specified interval. In this case, $\frac{9}{4}$ is in $(0,3)$.

Summary

We found the derivative of $g(x)$ using the chain rule. Using the given information that $g(x)$ has a local minimum at $x = \alpha$, we set $g'(\alpha) = 0$. Since $f''(x) > 0$, we know that $f'(x)$ is an increasing function, and therefore we could equate the arguments of $f'$ in the equation $f'\left(\frac{\alpha}{3}\right) = f'(3-\alpha)$. Solving for $\alpha$ and then calculating $8\alpha$, we found the answer to be 18.

Final Answer

The final answer is \boxed{18}, which corresponds to option (C).