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how to know the gradient of the curve y=2x^3-6x^2+8x+1 is never less than2?

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SupremeCookie said: The gradient of the curve is $\frac{dy}{dx}$. Find it. The gradient of the curve is minimized when $\frac{d^2y}{dx^2}=0$ Click to expand... Alternatively, Spoiler $\frac{dy}{dx}=6x^2-12x+8$ a parabola that concaves up, thus the minimum is at the vertex $\frac{-b}{2a}=\frac{-(-12)}{2\times6}=1$ and $\left . \dfrac{dy}{dx}\right |_{x=1} = 2$ Therefore, $\frac{dy}{dx} \ge 2$
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Spoiler $\dfrac{d^2 y}{dx^2} = 12x - 12\\~\\ \dfrac{dy}{dx} = 6x^2 - 12x + C\\~\\ \left . \dfrac{dy}{dx}\right |_{(2,9)} = 8\\~\\ 6(2^2)-12(2) + C = 8\\~\\ C=8\\~\\ \dfrac{dy}{dx} = 6x^2 - 12x + 8\\~\\ \text{As$\dfrac{d^3y}{dy^3}=12 > 0$, The minimum of$\dfrac{dy}{dx}$occurs at$\dfrac{d^2y}{dx^2}=12x-12=0$i.e.$x=1$}\\~\\ \left . \dfrac{dy}{dx} \right |_{x=1} = 6-12+8 = 2\\~\\ \text{and thus$\dfrac{dy}{dx} \geq 2$}$
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The gradient of the curve is $\frac{dy}{dx}$. Find it. The gradient of the curve is minimized when $\frac{d^2y}{dx^2}=0$
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