Maximum slope of the curve
Question:

Maximum slope of the curve $y=-x^{3}+3 x^{2}+9 x-27$ is

(a) 0

(b) 12

(c) 16

(d) 32

Solution:

The given curve is $y=-x^{3}+3 x^{2}+9 x-27$.

Slope of the curve, $m=\frac{d y}{d x}$

$\therefore m=\frac{d y}{d x}=-3 x^{2}+6 x+9$

$\Rightarrow \frac{d m}{d x}=-6 x+6$

For maxima or minima,

$\frac{d m}{d x}=0$

$\Rightarrow-6 x+6=0$

$\Rightarrow x=1$

Now,

$\frac{d^{2} m}{d x^{2}}=-6<0$

⇒ x  = 1 is the point of local maximum

So, the slope of the given curve is maximum when x = 1.

∴ Maximum value of the slope, m

$=-3 \times(1)^{2}+6 \times 1+9 \quad\left(m=-3 x^{2}+6 x+9\right)$

$=-3+6+9$

$=12$

Thus, the maximum slope of the curve $y=-x^{3}+3 x^{2}+9 x-27$ is 12 .

Hence, the correct answer is option (b).