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:

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 \(x=2\).

Answer 1

The Correct Option is A

Step 1: Differentiate the function.
For $f(x)=\dfrac{x}{2}+\dfrac{2}{x}$, \[ f'(x)=\frac{1}{2}-\frac{2}{x^2}. \]
Step 2: Find critical points.
Set $f'(x)=0$: $\dfrac12=\dfrac{2}{x^2}$, so $x^2=4$, giving $x=2$ or $x=-2$.
Step 3: Get the second derivative.
\[ f''(x)=\frac{4}{x^3}. \]
Step 4: Test $x=2$.
$f''(2)=\dfrac{4}{8}=\dfrac12>0$, so $x=2$ is a local minimum.
Step 5: Test $x=-2$.
$f''(-2)=\dfrac{4}{-8}=-\dfrac12<0$, so $x=-2$ is a local maximum.
Step 6: Conclude.
The local minimum is therefore at $x=2$.
\[ \boxed{x=2} \]