anonymous
  • anonymous
The function F(x)=e^(-3x^2) has 2 inflection points. Find the x-coordinate of the inflection point when x>0, correct to 2 decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
King, this is the same problem TheFigure was working on.... he just couldn't figure it out. (pun not intended lol)
anonymous
  • anonymous
@amistre64
KingGeorge
  • KingGeorge
The method he was using is definitely what you want to use. He probably just took one of the derivatives wrong.

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anonymous
  • anonymous
could u go see where he went wrong
KingGeorge
  • KingGeorge
Can you check if a certain answer is correct, and if so I'll tell you how to do it?
KingGeorge
  • KingGeorge
Try\[{1 \over \sqrt{6} }\approx 0.41\]
anonymous
  • anonymous
thats correct!!
KingGeorge
  • KingGeorge
After looking through Figure's work, I'm pretty sure he took the first derivative wrong. When he wrote\[f'(x)= -6x^2e^{-3x^2}\]It should have been\[f'(x)=-6xe^{-3x^2}\]Since the derivative of \(-3x^2\) is \(-6x\). Now, if we use this new \(f'\), we have that \[f''(x)=6e^{-3x^2}(6x^2-1)\]Thus, \(6x^2-1=0\) which means that \(x=\pm1/\sqrt{6}\) But since we're interested in positive x-values, \(x=1/\sqrt{6}\)
anonymous
  • anonymous
ok thanks for your time man. much appreciated.
KingGeorge
  • KingGeorge
You're welcome. Once again, sorry I wasn't able to answer this sooner.
anonymous
  • anonymous
at least you helped. thats all that matters.

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