Key Concepts and Formulas

• Finding Extrema using Derivatives: If a function $f(x)$ is continuous on a closed interval $[a, b]$ and differentiable on $(a, b)$, then its absolute maximum and minimum values occur either at the critical points (where $f'(x) = 0$ or is undefined) or at the endpoints of the interval.
• Chain Rule: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$
• Power Rule: $\frac{d}{dx} (x^n) = n x^{n-1}$

Step-by-Step Solution

Step 1: Determine the Domain of the Function

The function is given by $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$. For the function to be real-valued, the expressions inside the square roots must be non-negative.

• $x-2 \ge 0 \Rightarrow x \ge 2$
• $4-x \ge 0 \Rightarrow x \le 4$

Therefore, the domain of $f(x)$ is $[2, 4]$. This step is crucial because we will need to check the endpoints of this interval when finding the minimum and maximum values.

Step 2: Find the Derivative of the Function

We need to find $f'(x)$ to determine the critical points. Using the chain rule and power rule: $f'(x) = 3 \cdot \frac{1}{2\sqrt{x-2}} - \frac{1}{2\sqrt{4-x}} = \frac{3}{2\sqrt{x-2}} - \frac{1}{2\sqrt{4-x}}$

Step 3: Find the Critical Points

To find the critical points, we set $f'(x) = 0$ and solve for $x$: $\frac{3}{2\sqrt{x-2}} = \frac{1}{2\sqrt{4-x}}$ $3\sqrt{4-x} = \sqrt{x-2}$ Squaring both sides: $9(4-x) = x-2$ $36 - 9x = x - 2$ $38 = 10x$ $x = \frac{38}{10} = \frac{19}{5} = 3.8$

Since $2 \le 3.8 \le 4$, the critical point $x = \frac{19}{5}$ lies within the domain.

Step 4: Evaluate the Function at the Endpoints and Critical Point

We need to evaluate $f(x)$ at $x=2$, $x=4$, and $x=\frac{19}{5}$ to find the minimum and maximum values.

• $f(2) = 3\sqrt{2-2} + \sqrt{4-2} = 3(0) + \sqrt{2} = \sqrt{2}$
• $f(4) = 3\sqrt{4-2} + \sqrt{4-4} = 3\sqrt{2} + 0 = 3\sqrt{2}$
• $f\left(\frac{19}{5}\right) = 3\sqrt{\frac{19}{5}-2} + \sqrt{4-\frac{19}{5}} = 3\sqrt{\frac{9}{5}} + \sqrt{\frac{1}{5}} = 3\left(\frac{3}{\sqrt{5}}\right) + \frac{1}{\sqrt{5}} = \frac{9}{\sqrt{5}} + \frac{1}{\sqrt{5}} = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$

Step 5: Determine the Minimum and Maximum Values

We have $f(2) = \sqrt{2}$, $f(4) = 3\sqrt{2}$, and $f\left(\frac{19}{5}\right) = 2\sqrt{5}$. Since $\sqrt{2} \approx 1.414$, $3\sqrt{2} \approx 4.242$, and $2\sqrt{5} \approx 4.472$, we can conclude that:

• The minimum value is $\alpha = \sqrt{2}$
• The maximum value is $\beta = 2\sqrt{5}$

Step 6: Calculate $\alpha^2 + 2\beta^2$

We are asked to find $\alpha^2 + 2\beta^2$. $\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(2\sqrt{5})^2 = 2 + 2(4 \cdot 5) = 2 + 2(20) = 2 + 40 = 42$

Common Mistakes & Tips

• Forgetting to Check Endpoints: A common mistake is to only find the critical points and not check the endpoints of the interval. The maximum or minimum value could occur at an endpoint.
• Incorrect Differentiation: Be careful when differentiating the function, especially with the chain rule and square roots.
• Domain Restrictions: Always determine the domain of the function before proceeding. This will prevent you from considering values outside the domain.

Summary

We found the minimum and maximum values of the function $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$ by first determining its domain, then finding its derivative and critical points. We evaluated the function at the endpoints of the domain and at the critical point to find the minimum and maximum values, which were $\alpha = \sqrt{2}$ and $\beta = 2\sqrt{5}$ respectively. Finally, we calculated $\alpha^2 + 2\beta^2 = 42$.

Final Answer

The final answer is $\boxed{42}$, which corresponds to option (A).