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anonymous

  • 5 years ago

locate the three inflexion points of the curve y=xe^(-x^2). please show explanation :)

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  1. anonymous
    • 5 years ago
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    \[y=x*e^{-x^2}\]\[y'=e^{-x^2}(1-2x^2)\]\[y''=2e^{-x^2}x(2x^2-3)\] \[y''=0 \rightarrow x=\pm \sqrt{3/2}\]

  2. anonymous
    • 5 years ago
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    and x = 0

  3. anonymous
    • 5 years ago
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    When i diffrentiated y=x∗e−x2 i got y′=e−x2(1+2x2) can you show me how u got the first derivative

  4. anonymous
    • 5 years ago
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    use the product rule on x and e^(-x^2) and then the chain rule on e^(-x^2)

  5. anonymous
    • 5 years ago
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    i think you misused a sign somewhere for the quantity (1 + 2x^2)

  6. anonymous
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    e should be -e

  8. anonymous
    • 5 years ago
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    v' is -2x e^(-x^2)

  9. anonymous
    • 5 years ago
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    oh yes i see that mistake thanks heaps

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