Question

The function \( f(x) = \frac{x}{2} + \frac{2}{x} \) has a local minimum at:

• A. \( x = 2 \)
• B. \( x = -2 \)
• C. \( x = 0 \)
• D. \( x = 1 \)

Hint:

To determine the local minima or maxima of a function, find the critical points by setting the first derivative equal to zero, and then use the second derivative test to classify the points. A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.

Answer 1

The Correct Option is A

We are given the function: \[ f(x) = \frac{x}{2} + \frac{2}{x} \] We need to find where this function has a local minimum. Step 1: Find the first derivative of \( f(x) \). To find the critical points, we first take the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( \frac{x}{2} + \frac{2}{x} \right) \] Using the power rule for derivatives: \[ f'(x) = \frac{1}{2} - \frac{2}{x^2} \] Step 2: Set the first derivative equal to zero to find critical points. To find the critical points, set \( f'(x) = 0 \): \[ \frac{1}{2} - \frac{2}{x^2} = 0 \] Solving for \( x \): \[ \frac{2}{x^2} = \frac{1}{2} \] Multiplying both sides by \( x^2 \) and simplifying: \[ 2 = \frac{x^2}{2} \] \[ x^2 = 4 \] \[ x = \pm 2 \] Thus, the critical points are \( x = 2 \) and \( x = -2 \). Step 3: Determine the nature of the critical points using the second derivative. To determine whether these critical points correspond to local minima or maxima, we take the second derivative of \( f(x) \): \[ f''(x) = \frac{d}{dx} \left( \frac{1}{2} - \frac{2}{x^2} \right) \] Using the power rule again: \[ f''(x) = \frac{4}{x^3} \] Now, evaluate \( f''(x) \) at \( x = 2 \) and \( x = -2 \): - At \( x = 2 \), \( f''(B) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} \), which is positive, indicating a local minimum at \( x = 2 \). - At \( x = -2 \), \( f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2} \), which is negative, indicating a local maximum at \( x = -2 \). Step 4: Conclusion. Since \( x = 2 \) corresponds to a local minimum, the correct answer is option (A)