The maximum slope of the curve y=-x^{3}+3x^{2 | vedclass

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Question

(C) Given the curve $y = -x^{3} + 3x^{2} + 2x - 27$. To find the slope of the curve,we differentiate with respect to $x$: $\frac{dy}{dx} = -3x^{2} + 6

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the curve $y = -x^{3} + 3x^{2} + 2x - 27$. To find the slope of the curve,we differentiate with respect to $x$: $\frac{dy}{dx} = -3x^{2} + 6x + 2$. Let the slope be $m = -3x^{2} + 6x + 2$. To find the maximum slope,we differentiate $m$ with respect to $x$: $\frac{dm}{dx} = -6x + 6$. Setting $\frac{dm}{dx} = 0$ for critical points: $-6x + 6 = 0 \implies x = 1$. Checking the second derivative: $\frac{d^{2}m}{dx^{2}} = -6 Since the second derivative is negative,the slope is maximum at $x = 1$. Substituting $x = 1$ into the expression for $m$: $m_{	ext{max}} = -3(1)^{2} + 6(1) + 2 = -3 + 6 + 2 = 5$.

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

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