anonymous
  • anonymous
what are the inflection points f(x)=e^(-x^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Find its second derivative and make it equal to zero and solve for x
zepdrix
  • zepdrix
Look back at your previous question ames, those were your inflection points. :) The x='s.
anonymous
  • anonymous
i could have sworn we just did this one

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anonymous
  • anonymous
thats what i thought but then how do i tell if it concaved up or concaved down?
anonymous
  • anonymous
second derivative is \[f''(x)=(4x^2-2)e^{-x^2}\] which is zero if \[4x^2-2=0\] or \[x=\pm\frac{1}{\sqrt2}\]
anonymous
  • anonymous
concave up if \(f''(x)>0\)
anonymous
  • anonymous
ignore the \(e^{-x^2}\) part as it is always posive solve \(4x^2-2>0\)
anonymous
  • anonymous
\[y=4x^2-2\] is a parabola that faces up, so it is negative between the zeros and positive outside of them
anonymous
  • anonymous
i.e. \(f''(x)<0\) on the interval \((-\frac{1}{\sqrt2}, \frac{1}{\sqrt2})\)
anonymous
  • anonymous
positive otherwise this more or less clear?
anonymous
  • anonymous
yes thank you.. i got a bit confused

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