Explanation

(b)
Let $f(x) = - x^{3} + 3x^{2} + 9x - 27$
The slope of this curve $f^{'}(x) = - 3x^{2} + 6x + 9$
Let $g(x) = f^{'}(x) = - 3x^{2} + 6x + 9$
On differentiating w.r.t. $x$, we get
$g^{'}(x) = - 6x + 6$
For maxima or minima put $g^{'}(x) = 0\ \Rightarrow x = 1$
Now, $g^{''}(x) = - 6 < 0$ and hence, at $x = 1,\ \ g(x)$ (slope) will have maximum value
$\therefore\ \ \ \left\lbrack g(1) \right\rbrack_{\max} = - 3 \times 1 + 6(1) + 9 = 12$