Explanation

(b)
Given, $f(x) = x + \frac{1}{x}$
$\Rightarrow \ \ \ \ \ \ \ f^{'}(x) = 1 - \frac{1}{x^{2}}$
$\Rightarrow$ For local maxima or local minima, put
$f^{'}(x) = 0$
$\Rightarrow \ \ \ \ \ \ 1 - \frac{1}{x^{2}} = 0$
$\Rightarrow x = \pm 1$
Now, $f^{''}(x) = \frac{2}{x^{3}}$
At $x = 1$, $f^{''}(x) = \frac{2}{1^{3}} > 0$, local minima
At $x = - 1,\ \ f’’(x) = \frac{2}{( - 1)^{3}}$
$= - 2 < 0$, local maxima