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How would I figure out the following question.

Find the values of x at which the rate of change of $y=30+28x^2+16x^3-2x^4$ with respect to $x$ is zero.

Do I have to take the derivative and set it to zero or something else.

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Yes ${}{}{}{}{}{}{}{}$ –  Amr Feb 10 '13 at 17:44
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Yes, the rate of change is the derivative, so you gotta set the derivative equal to zero. –  ciceksiz kakarot Feb 10 '13 at 17:44
    
Hmm but how many times would I take derivative only one time or several times. –  Fernando Martinez Feb 10 '13 at 17:47
    
Just like ciceksizkakarot said ... –  Hagen von Eitzen Feb 10 '13 at 17:54
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1 Answer

up vote 2 down vote accepted

The first derivative is the same thing as the rate of change, so the question essentially says to find when the first derivative equals zero. So you have $$ \frac {dy} {dx} = -8x(x-7)(x+1) = 0, $$ so the x-coordinates where the rate of change is zero are $x = 0, 7,$ and $-1$.

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Thanks Von Eitzen –  Fernando Martinez Feb 10 '13 at 18:04
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