The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minima at $x$ equal to: |
$2$ $1$ $0$ $-2$ |
$2$ |
The correct answer is Option (1) → $2$ ## We have, $f(x) = \frac{x}{2} + \frac{2}{x}$ $f'(x) = \frac{1}{2} - \frac{2}{x^2}$ Put, $f'(x) = 0 \Rightarrow \frac{1}{2} - \frac{2}{x^2} = 0$ $\Rightarrow \frac{1}{2} = \frac{2}{x^2}$ $\Rightarrow x^2 = 4$ $\Rightarrow x = \pm 2$ Now, $f''(x) = \frac{4}{x^3}$ At $x = 2$, $f''(2) = \frac{4}{8} = \frac{1}{2} > 0$ At $x = -2$, $f''(-2) = \frac{4}{-8} = -\frac{1}{2} < 0$ Thus, $f(x)$ has local minima at $x = 2$. |