The points on the curve y = 12x - x^3 at whic | vedclass

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(D) Given the curve $y = 12x - x^3$. To find the points where the gradient is zero,we calculate the derivative $\frac{dy}{dx}$ and set it to $0$. $\fr

Vedclass · Accepted Answer

$(2, 16), (-2, -16)$

Vedclass · Answer

(D) Given the curve $y = 12x - x^3$. To find the points where the gradient is zero,we calculate the derivative $\frac{dy}{dx}$ and set it to $0$. $\frac{dy}{dx} = \frac{d}{dx}(12x - x^3) = 12 - 3x^2$. Setting the gradient to zero: $12 - 3x^2 = 0$. $3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. For $x = 2$,$y = 12(2) - (2)^3 = 24 - 8 = 16$. So,the point is $(2, 16)$. For $x = -2$,$y = 12(-2) - (-2)^3 = -24 - (-8) = -24 + 8 = -16$. So,the point is $(-2, -16)$. Thus,the points are $(2, 16)$ and $(-2, -16)$.

The points on the curve $y = 12x - x^3$ at which the gradient is zero are

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