Question:
The function \(f(x) = \frac{x}{2} + \frac{2}{x}\) has a local minimum at
The function \(f(x) = \frac{x}{2} + \frac{2}{x}\) has a local minimum at
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For any positive \(x\), the Arithmetic Mean-Geometric Mean (AM-GM) inequality states that \(\frac{x}{2} + \frac{2}{x} \ge 2\sqrt{\frac{x}{2} \cdot \frac{2}{x}} = 2\). The minimum value of 2 occurs when \(\frac{x}{2} = \frac{2}{x}\), which confirms \(x=2\).
Updated On: Jun 9, 2026
- \(x = 2\)
- \(x = -2\)
- \(x = 0\)
- \(x = 1\)
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The Correct Option is A
Solution and Explanation
Concept:
To find the local extrema of a function \(f(x)\), we calculate the first derivative \(f'(x)\), set it to zero to find the critical points, and then use the second derivative test \(f''(x)\) to determine if the point is a local maximum or minimum.
Step 1: Find the first derivative \(f'(x)\).
Given the function: \[ f(x) = \frac{x}{2} + \frac{2}{x} \] Differentiating with respect to \(x\): \[ f'(x) = \frac{1}{2} - \frac{2}{x^2} \]
Step 2: Find the critical points by setting \(f'(x) = 0\).
\[ \frac{1}{2} - \frac{2}{x^2} = 0 \] \[ \frac{1}{2} = \frac{2}{x^2} \] \[ x^2 = 4 \implies x = 2, \quad x = -2 \]
Step 3: Apply the second derivative test.
Calculate the second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{1}{2} - 2x^{-2} \right) = 0 - 2(-2)x^{-3} = \frac{4}{x^3} \] Test \(x = 2\): \[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = 0.5 \] Since \(f''(2) > 0\), the function has a local minimum at \(x = 2\). Test \(x = -2\): \[ f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -0.5 \] Since \(f''(-2) < 0\), the function has a local maximum at \(x = -2\). center minipage0.3
Local minimum at x = 2 minipage center
Step 1: Find the first derivative \(f'(x)\).
Given the function: \[ f(x) = \frac{x}{2} + \frac{2}{x} \] Differentiating with respect to \(x\): \[ f'(x) = \frac{1}{2} - \frac{2}{x^2} \]
Step 2: Find the critical points by setting \(f'(x) = 0\).
\[ \frac{1}{2} - \frac{2}{x^2} = 0 \] \[ \frac{1}{2} = \frac{2}{x^2} \] \[ x^2 = 4 \implies x = 2, \quad x = -2 \]
Step 3: Apply the second derivative test.
Calculate the second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{1}{2} - 2x^{-2} \right) = 0 - 2(-2)x^{-3} = \frac{4}{x^3} \] Test \(x = 2\): \[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = 0.5 \] Since \(f''(2) > 0\), the function has a local minimum at \(x = 2\). Test \(x = -2\): \[ f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -0.5 \] Since \(f''(-2) < 0\), the function has a local maximum at \(x = -2\). center minipage0.3
Local minimum at x = 2 minipage center
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