Explanation

(a)
$\because\ \ \ f(x) = 2x^{3} - 15x^{2} + 36x + 4$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 6x^{2} - 30x + 36$ …(i)
For maxima or minima $f^{'}(x) = 0$
$\Rightarrow \ \ \ 6x^{2} - 30x + 36 = 0$
$\Rightarrow \ \ x^{2} - 5x + 6 = 0$
$\Rightarrow \ \ \ (x - 2)(x - 3) = 0\ \ \ \Rightarrow \ \ \ x = 2,\ 3$
Again differentiating Eq. (i), we get
$f^{''}(x) = 12x - 30$
$\Rightarrow \ \ \ f^{''}(2)\text{=}24 - 30 = - 6 < 0$
Therefore, $f(x)$ is maximum at, $x = 2$