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anonymous

  • 5 years ago

I can't find critical points for y=x(9-x^2)^1/2

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  1. anonymous
    • 5 years ago
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    First take the derivative dy/dx=\[(9-2x ^{2})/\sqrt{9-x ^{2}}\]

  2. anonymous
    • 5 years ago
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    next we must find where dy/dx = 0 or where it is undefined

  3. anonymous
    • 5 years ago
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    dy/dx = 0 for \[x = \pm \sqrt{9/2}\]

  4. anonymous
    • 5 years ago
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    dy/dx is undefined for \[x = \pm3\], when the denominator equals 0

  5. anonymous
    • 5 years ago
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    Are you looking for a specific type of critical point?

  6. anonymous
    • 5 years ago
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    \[y = x \sqrt{9-x^2}\] the derivative of it is: \[y'=\sqrt{9-x^2} -x^2/\sqrt{9-x^2}\] then y'=0 finely x =2.45

  7. anonymous
    • 5 years ago
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    that is not the only answer.

  8. anonymous
    • 5 years ago
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    and the derivative is not quite correct.

  9. anonymous
    • 5 years ago
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    The critical points should be \[x = \pm \sqrt{9/2}\]

  10. anonymous
    • 5 years ago
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    oops, sorry corec, the derivative is fine

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