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Step 1: Find the first derivative of the function.
Given \(f(x) = \frac{x}{2} + \frac{2}{x}\), we find its derivative \(f'(x)\):
\(f'(x) = \frac{1}{2} - \frac{2}{x^2}\)
Step 2: Find the critical points by setting the first derivative equal to zero.
\(\frac{1}{2} - \frac{2}{x^2} = 0\)
\(\frac{1}{2} = \frac{2}{x^2}\)
\(x^2 = 4\)
\(x = \pm 2\)
Step 3: Find the second derivative of the function.
\(f''(x) = \frac{d}{dx} (\frac{1}{2} - \frac{2}{x^2}) = \frac{4}{x^3}\)
Step 4: Apply the second derivative test to determine local minima.
For \(x = 2\), \(f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0\). Since the second derivative is positive, \(x = 2\) is a local minima.
For \(x = -2\), \(f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2} < 0\). Since the second derivative is negative, \(x = -2\) is a local maxima.
Step 5: Identify the x value where the function has a local minima.
The function has a local minima at \(x = 2\).
Correct Answer: 2