Explanation

(b)
Given function is $f(x) = x + \sin x$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 1 + \cos x$
For maxima or minima put $f^{'}(x) = 0$
$\Rightarrow \ \ 1 + \cos x = 0\ \ \ \Rightarrow \ \cos x = - 1\ \ \ \Rightarrow \ \ \ x = \pi$
Again differentiating w.r.t. $x$, we get
$f^{''}(x) = - \sin x,\ at\ \ x = \pi f^{''}(\pi) = 0$
Again differentiating w.r.t. $x$, we get $f^{'''}(x) = - \cos x$,
$f^{'''}(\pi) = 1$
At $x = \pi,\ f(x)$ is minimum