how can i find the coordinates of the turning point 4x(x-1)(x-2)?anyone can help.
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Originally Posted by mastermin346 how can i find the coordinates of the turning point 4x(x-1)(x-2)?anyone can help. $\displaystyle y=4x(x-1)(x-2)=4(x^3-3x^2+2x)$ $\displaystyle \frac{dy}{dx}=4(3x^2-6x+2)$ At the turning point, $\displaystyle \frac{dy}{dx}=0$ $\displaystyle x=1\pm{\frac{1}{\sqrt{3}}}$
Last edited by alexmahone; Oct 13th 2009 at 11:46 PM.
thank you
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