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

1

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3

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\frac{1}{2}

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- \frac{1}{2}

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Solution

The correct option is A $1$
Given, $y = - x^{3} + 3 x^{2} + 9 x - 27$
Slope at any point on thegiven curve is $m = \frac{d y}{d x} = - 3 x^{2} + 6 x + 9$
Now for maximum value of slope we must have $\frac{d m}{d x} = 0 = - 6 x + 6 \Rightarrow x = 1$

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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.

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

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