Key Concepts and Formulas

• First Derivative Test: Critical points of a function $f(x)$ occur where $f'(x) = 0$ or $f'(x)$ is undefined. These are potential locations of local maxima or minima.
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f(x)$ has a local maximum at $x=c$. If $f''(c) = 0$, the test is inconclusive.
• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$

Step-by-Step Solution

Step 1: Analyze the given function and its domain. The given function is $f(x) = \frac{x}{2} + \frac{2}{x}$. It's important to note the domain of the function. Since division by zero is undefined, $x \neq 0$. This constraint must be respected throughout our calculations.

Step 2: Calculate the first derivative, $f'(x)$. To find the critical points, we need to find the first derivative. We can rewrite the function as $f(x) = \frac{1}{2}x + 2x^{-1}$. Now, differentiate using the power rule: $f'(x) = \frac{d}{dx}\left(\frac{1}{2}x\right) + \frac{d}{dx}\left(2x^{-1}\right)$ $f'(x) = \frac{1}{2}(1)x^{1-1} + 2(-1)x^{-1-1}$ $f'(x) = \frac{1}{2} - 2x^{-2}$ $f'(x) = \frac{1}{2} - \frac{2}{x^2}$ Explanation: We find the first derivative because critical points (where local extrema can occur) are found where $f'(x) = 0$ or where $f'(x)$ is undefined.

Step 3: Find the critical points. To find the critical points, we set $f'(x) = 0$: $\frac{1}{2} - \frac{2}{x^2} = 0$ Now, solve for $x$: $\frac{1}{2} = \frac{2}{x^2}$ Multiply both sides by $2x^2$ (since $x \neq 0$): $x^2 = 4$ Taking the square root of both sides gives: $x = \pm\sqrt{4}$ $x = 2 \quad \text{or} \quad x = -2$ These are our critical points. Both $x=2$ and $x=-2$ are within the domain of $f(x)$.

Step 4: Calculate the second derivative, $f''(x)$. To classify these critical points, we need to find the second derivative. We differentiate $f'(x) = \frac{1}{2} - 2x^{-2}$ with respect to $x$: $f''(x) = \frac{d}{dx}\left(\frac{1}{2}\right) - \frac{d}{dx}\left(2x^{-2}\right)$ $f''(x) = 0 - 2(-2)x^{-2-1}$ $f''(x) = 4x^{-3}$ $f''(x) = \frac{4}{x^3}$ Explanation: The second derivative helps us determine the concavity. If $f''(x) > 0$, the function is concave up (like a valley), indicating a local minimum. If $f''(x) < 0$, it's concave down (like a hill), indicating a local maximum.

Step 5: Apply the Second Derivative Test to classify critical points. Now, evaluate $f''(x)$ at each critical point:

• For $x=2$: $f''(2) = \frac{4}{(2)^3} = \frac{4}{8} = \frac{1}{2}$ Since $f''(2) = \frac{1}{2} > 0$, the function $f(x)$ has a local minimum at $x=2$.

• For $x=-2$: $f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2}$ Since $f''(-2) = -\frac{1}{2} < 0$, the function $f(x)$ has a local maximum at $x=-2$.

The question asks for the local minimum, which we found to be at $x=2$.

Common Mistakes & Tips:

• Domain Check: Always check the domain of the function. Critical points outside the domain are not valid.
• Sign Interpretation: Understand the meaning of the sign of the second derivative: positive means local minimum, negative means local maximum.

Summary

By applying the Second Derivative Test, we found the critical points where the derivative is zero. Then, by evaluating the second derivative at these points, we determined that $x=2$ corresponds to a local minimum.

The final answer is $\boxed{2}$, which corresponds to option (A).