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Find the value of $x$ where the tangent line to the curve $y=2xe^{-x}$ is horizontal. I need your help please...

Macavity
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Alice1997
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  • What have you done so far on the problem? – Macavity Oct 15 '13 at 11:06
  • Hello, welcome to Math.SE. Thank you for your question! It is good practice on this site to add a bit of information on the context your question came up in, and to share your own work on it. It's also fine if you state that you're completely lost -- the information is helpful for answerers to gauge their answer. For more information on asking a good question on this site, see here. – Lord_Farin Oct 15 '13 at 11:27

1 Answers1

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Hint: Why is it sufficient to set the derivative equal to $0$ and solve?

Clayton
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  • since the derivative is the slope of the tangent line, equating it to 0 means that there is no slope at all (horizontal line has no slope) – Alice1997 Oct 15 '13 at 11:08
  • i dont know how to simplify the derivative in the problem so i cant find the value of x – Alice1997 Oct 15 '13 at 11:09
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    @Alice1997 What is the derivative? Isn't it the sum of two terms? Do those terms have something in common? – Arthur Oct 15 '13 at 11:11
  • @Alice1997: Notice that $e^{-x}>0$ for all $x$... – Clayton Oct 15 '13 at 11:14
  • @Alice1997 So you're looking for the derivative of $y=2xe^{-x}$. –  Oct 15 '13 at 11:18
  • @Arthur i don't know... 2? – Alice1997 Oct 15 '13 at 11:47
  • Are you aware of the product rule? (this is what you need) @Alice – George Tomlinson Oct 15 '13 at 12:00
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    @bluesh34 yeah. i tried getting the derivative and this is what i got

    f'(x)=2x(e^-x)(-1)+0+2(1)(e^-x) =-(2xe^-x)+(2e^-x)

     – Alice1997 Oct 15 '13 at 12:23
  • Well your answer is correct, although you don't really need the 0 in your initial calculation: consider it as the product of $2x$ and $e^{-x}$. Anyway, the next step is to factorise your answer, so that it's easier to see which value of $x$ makes it 0. Can you factorise the expression? – George Tomlinson Oct 15 '13 at 17:59