anonymous
  • anonymous
I can't find critical points for y=x(9-x^2)^1/2
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
First take the derivative dy/dx=\[(9-2x ^{2})/\sqrt{9-x ^{2}}\]
anonymous
  • anonymous
next we must find where dy/dx = 0 or where it is undefined
anonymous
  • anonymous
dy/dx = 0 for \[x = \pm \sqrt{9/2}\]

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anonymous
  • anonymous
dy/dx is undefined for \[x = \pm3\], when the denominator equals 0
anonymous
  • anonymous
Are you looking for a specific type of critical point?
anonymous
  • anonymous
\[y = x \sqrt{9-x^2}\] the derivative of it is: \[y'=\sqrt{9-x^2} -x^2/\sqrt{9-x^2}\] then y'=0 finely x =2.45
anonymous
  • anonymous
that is not the only answer.
anonymous
  • anonymous
and the derivative is not quite correct.
anonymous
  • anonymous
The critical points should be \[x = \pm \sqrt{9/2}\]
anonymous
  • anonymous
oops, sorry corec, the derivative is fine

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