The function $f(x) = x + \frac{1}{x}$ has

local maxima at $x = -1$

The correct answer is Option (1) → local maxima at $x = -1$

Given $f(x) = x + \frac{1}{x}$.

Differentiate:

$f'(x) = 1 - \frac{1}{x^2}$

Setting $f'(x) = 0$:

$1 - \frac{1}{x^2} = 0$

$x^2 = 1$

$x = \pm 1$

Second derivative:

$f''(x) = \frac{2}{x^3}$

At $x = 1$: $f''(1) = \frac{2}{1^3} = 2 > 0$ (local minimum).

At $x = -1$: $f''(-1) = \frac{2}{(-1)^3} = -2 < 0$ (local maximum).

Local maxima occur at $x = -1$ with value:

$f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$.