Key Concepts and Formulas

• Finding Extrema Using Derivatives: To find local minima or maxima of a function $f(x)$, find the critical points by solving $f'(x) = 0$. The second derivative test can confirm if a critical point is a minimum ($f''(x) > 0$) or maximum ($f''(x) < 0$).
• AM-GM Inequality: For non-negative real numbers $a_1, a_2, ..., a_n$, we have $\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$, with equality when $a_1 = a_2 = ... = a_n$.

Step-by-Step Solution

Step 1: State the given function and its domain. We have $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$ for $x > 0$.

Step 2: Find the first derivative, $f'(x)$. We differentiate $f(x)$ with respect to $x$ to find the critical points: $f'(x) = \frac{d}{dx} \left( \frac{5x^2}{2} + \frac{\alpha}{x^5} \right) = \frac{d}{dx} \left( \frac{5}{2}x^2 + \alpha x^{-5} \right)$ Using the power rule: $f'(x) = \frac{5}{2}(2x) + \alpha(-5x^{-6}) = 5x - \frac{5\alpha}{x^6}$

Step 3: Find the critical points by setting $f'(x) = 0$. $5x - \frac{5\alpha}{x^6} = 0$ $5x = \frac{5\alpha}{x^6}$ $x = \frac{\alpha}{x^6}$ $x^7 = \alpha$ $x = \alpha^{1/7}$

Step 4: Find the second derivative, $f''(x)$. To confirm that $x = \alpha^{1/7}$ corresponds to a minimum, we find the second derivative: $f''(x) = \frac{d}{dx} \left( 5x - 5\alpha x^{-6} \right) = 5 - 5\alpha(-6x^{-7}) = 5 + 30\alpha x^{-7} = 5 + \frac{30\alpha}{x^7}$

Step 5: Verify it's a minimum using the second derivative test. We evaluate $f''(x)$ at the critical point $x = \alpha^{1/7}$: $f''(\alpha^{1/7}) = 5 + \frac{30\alpha}{(\alpha^{1/7})^7} = 5 + \frac{30\alpha}{\alpha} = 5 + 30 = 35$ Since $f''(\alpha^{1/7}) = 35 > 0$, the critical point corresponds to a minimum.

Step 6: Substitute the critical point into $f(x)$ and set it equal to the given minimum value. We are given that the minimum value is 14. So, $f(\alpha^{1/7}) = \frac{5(\alpha^{1/7})^2}{2} + \frac{\alpha}{(\alpha^{1/7})^5} = 14$ $\frac{5\alpha^{2/7}}{2} + \frac{\alpha}{\alpha^{5/7}} = 14$ $\frac{5\alpha^{2/7}}{2} + \alpha^{1 - 5/7} = 14$ $\frac{5\alpha^{2/7}}{2} + \alpha^{2/7} = 14$ $\alpha^{2/7} \left( \frac{5}{2} + 1 \right) = 14$ $\alpha^{2/7} \left( \frac{7}{2} \right) = 14$ $\alpha^{2/7} = 14 \cdot \frac{2}{7} = 4$ $\alpha^{2/7} = 4 = 2^2$ Raise both sides to the power of $\frac{7}{2}$: $\left( \alpha^{2/7} \right)^{7/2} = \left( 2^2 \right)^{7/2}$ $\alpha = 2^{2 \cdot (7/2)} = 2^7 = 128$

Step 7: Check if AM-GM is applicable and yields a simpler solution. We can apply AM-GM inequality to $\frac{5x^2}{2}$ and $\frac{\alpha}{x^5}$ directly if we can make the powers of $x$ cancel out when we take the product. Let's rewrite the function as a sum of 7 terms: $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5} = \frac{5x^2}{2} = \frac{5x^2}{2 \cdot 5/2}+\frac{5x^2}{2 \cdot 5/2} + \frac{5x^2}{2 \cdot 5/2} + \frac{5x^2}{2 \cdot 5/2}+\frac{5x^2}{2 \cdot 5/2}$ $f(x) = \frac{5x^2}{7/2} \frac{5x^2}{7/2} \frac{5x^2}{7/2} \frac{5x^2}{7/2} \frac{5x^2}{7/2}$ Instead, let's try to write $\frac{5x^2}{2}$ as a sum of terms such that the power of $x$ cancels out with $\frac{\alpha}{x^5}$ when we apply AM-GM. We want to write $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5} = \frac{ax^2}{2} + \frac{ax^2}{2} + ... + \frac{ax^2}{2} + \frac{\alpha}{x^5}$ such that $2k-5=0$, where k is the number of times $\frac{ax^2}{2}$ is added. This is not possible with integers.

However, consider rewriting the first term: $\frac{5x^2}{2} = \frac{5}{7/2} \frac{5x^2}{5/7}$ Then use AM-GM on the terms \frac{\frac{5x^2}{7/5} \frac{2}{5x^2}{5/7} \frac{2}{5x^2}{5/7} \frac{2}{5x^2}{5/7} \frac{2}{5x^2}{5/7} \frac{2}{5x^2}{5/7}\frac{2}{5x^2}{5/7}

Let's go back to derivatives. We found $f(\alpha^{1/7}) = 14$, and $\alpha = 128$.

Common Mistakes & Tips

• Always check the second derivative to confirm whether a critical point is a minimum or maximum.
• The AM-GM inequality can be a powerful tool, but it requires careful manipulation of the function to ensure that the product of the terms is constant or easily simplified.
• Be careful with algebraic manipulations and exponents, especially when taking roots.

Summary

We found the critical point of the function $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$ by setting its derivative equal to zero. We then used the second derivative test to confirm that this critical point corresponds to a minimum. Substituting the critical point back into the original function and equating it to the given minimum value of 14 allowed us to solve for $\alpha$, yielding $\alpha = 128$.

Final Answer

The final answer is \boxed{128}, which corresponds to option (C).