Explanation

(b)
Let $f(x) = x + \frac{1}{x}\ \ \ \Rightarrow \ \ \ f^{'}(x) = 1 - \frac{1}{x^{2}}$
For maxima and minima, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ \ 1 - \frac{1}{x^{2}} = 0\ \ \ \ \Rightarrow \ \ \ x = \pm 1$
Now, $f^{''}(x) = \frac{2}{x^{3}}$
At $x = 1,\ \ \ \ f^{''}(x) = \frac{2}{x^{3}}$
At $x = 1$, $f^{''}(x) = +$ ve, minima
And at $x = - 1,\ \ f^{''}(x) = -$ve, maxima
Thus, $f(x)$ attains minimum value at $x = 1$