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I need help with the following question.

Find the largest positive value of x at which the curve:

$y = (2x + 7)^6 (x - 2)^5$

has a horizontal tangent line.

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    How do I proceed after finding the derivative of y?2012-03-19

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To solve y'(x)=0 when $y(x)\ne0$, one can consider the derivative \dfrac{y'(x)}{y(x)} of the function $\log|y(x)|=6\log|2x+7|+5\log|x-2|$. The computations become much simpler since \frac{y'(x)}{y(x)}=\frac{6\cdot2}{2x+7}+\frac5{x-2}. Thus, y'(x)=0 as soon as the RHS is zero, that is, when $12(x-2)+5(2x+7)=0$, that is, when $x=-\frac12$. Complete the reasoning with the values where $y(x)=0$, that is, $x=-\frac72$ and $x=2$.

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    @Didier Can I find you in the chat?2012-03-19
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Hint: what slope corresponds to a horizontal tangent? How do you find the slope of a tangent line?

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    @Xavier: Yes, you are on the right track. As Kirthi says, you want the derivative to be zero, as that is the slope of the tangent.2012-03-19
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Hint: $\displaystyle{\frac{dy}{dx} = \left(12(2x+7)^5(x-2)^5+5(2x+7)^6(x-2)^4 \right) = 0}$ , in other words

$\displaystyle{\frac{dy}{dx} = (2x+7)^5(x-2)^4(22x+11) = 0}$ at what points?

$\displaystyle{\frac{dy}{dx} = 0}$ at $x=-\frac{7}{2}, x=2, x=-\frac{1}{2}$

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    @Xavier, corrected that but the approach is still the same. (I just gave a hint)2012-03-19