anonymous
  • anonymous
locate the three inflexion points of the curve y=xe^(-x^2). please show explanation :)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[y=x*e^{-x^2}\]\[y'=e^{-x^2}(1-2x^2)\]\[y''=2e^{-x^2}x(2x^2-3)\] \[y''=0 \rightarrow x=\pm \sqrt{3/2}\]
anonymous
  • anonymous
and x = 0
anonymous
  • anonymous
When i diffrentiated y=x∗e−x2 i got y′=e−x2(1+2x2) can you show me how u got the first derivative

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anonymous
  • anonymous
use the product rule on x and e^(-x^2) and then the chain rule on e^(-x^2)
anonymous
  • anonymous
i think you misused a sign somewhere for the quantity (1 + 2x^2)
anonymous
  • anonymous
i dont see where i went wrong heres my working out y'= uv'+vu' where u= x u'= 1 v=e^(-x^2) v'=2xe^(-x^2) then y'= x(2xe^(-x^2))+e^(-x^2)(1) = e^(-x^2)(2x^2+1) can u tell me where my mistake is
anonymous
  • anonymous
e should be -e
anonymous
  • anonymous
v' is -2x e^(-x^2)
anonymous
  • anonymous
oh yes i see that mistake thanks heaps

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