# Find the largest positive value of $x$ at which the curve $y = (2x + 7)^6 (x - 2)^5$ has a horizontal tangent line.

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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Hint: Horizontal tangent line - is $\frac{dy}{dx} 0,$ or positive or negative? – Siddhi V Iyer Mar 19 '12 at 20:49
A sketch often helps - easy to do here by first sketching the sextic and quintic factors. – Mark Bennet Mar 19 '12 at 20:58
How do I proceed after finding the derivative of y? – Xavier Mar 19 '12 at 21:37

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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Thanks for the insight, really makes the computation much simpler. – Xavier Mar 19 '12 at 22:15
@Didier Can I find you in the chat? – Pedro Tamaroff Mar 19 '12 at 22:17

Hint: what slope corresponds to a horizontal tangent? How do you find the slope of a tangent line?

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Am I on the right track to first find the derivative of y? What do I do next? – Xavier Mar 19 '12 at 20:56
@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. – Ross Millikan Mar 19 '12 at 21:44

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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Hi, shouldn't the derivative be 12(2x+7)^5(x-2)^5 + 5(x-2)^4(2x+7)^6 ? Taking into account of chain-rule. Do correct me if I am wrong. – Xavier Mar 19 '12 at 21:20
@Xavier: You are correct. – Joe Johnson 126 Mar 19 '12 at 21:33
@JoeJohnson126: Thanks. Do you have any idea how to proceed from there? – Xavier Mar 19 '12 at 21:36
@Xavier, corrected that but the approach is still the same. (I just gave a hint) – Kirthi Raman Mar 19 '12 at 21:44