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

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Solution

The correct option is B 12
Slope $m = \frac{d y}{d x} = - 3 x^{2} + 6 x + 9$
$m^{'} (x) = - 6 x + 6$
For maxima or minima,
$m^{'} (x) = 0$
$\Rightarrow x = 1$
$m^{''} (x) = - 6 < 0$
Hence slope is maximum at $x = 1$
$m = 12$

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y = - x^{3} + 3 x^{2} + 9 x - 27

y = - x^{3} + 3 x^{2} + 9 x - 27

At what point is the slope of the curve $y = - x^{3} + 3 x^{2} + 9 x - 27$ maximum? Also, find the maximum slope.

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