Maximum slope of the curve $y=-x^{3}+3 x^{2}+9 x-27$ is
(a) 0
(b) 12
(c) 16
(d) 32
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).
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