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Math Help - Point of Inflection (second derivative)- easy

  1. #1
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    Post Point of Inflection (second derivative)- easy

    Can anyone solve this one?
    I got stuck..

    f(x) = 4/ (x^2 + 8)

    I need f''(x) second derivative for this
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  2. #2
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    Quote Originally Posted by dkssudgktpdy View Post
    Can anyone solve this one?
    I got stuck..

    f(x) = 4/ (x^2 + 8)

    I need f''(x) second derivative for this
    Can you get the first derivative? Please show all your work.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Can you get the first derivative? Please show all your work.

    yup.
    f(x) = 4 (x^2 +8)^-1

    = -4 (x^2 +8)^-2 (2x)

    = -8x (x^2 +8)^-2
    f''(x)= 16x (x^2 +8)^-3 (2x)
    = 32x (x^2 +8)^-3
    0 = 32x/ (x^2 +8)^3
    x=0

    and.. 0 is not the right answer. Where did I make a mistake?
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  4. #4
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    Quote Originally Posted by dkssudgktpdy View Post
    yup.
    f(x) = 4 (x^2 +8)^-1

    = -4 (x^2 +8)^-2 (2x)

    = -8x (x^2 +8)^-2
    f''(x)= 16x (x^2 +8)^-3 (2x)
    = 32x (x^2 +8)^-3
    0 = 32x/ (x^2 +8)^3
    x=0

    and.. 0 is not the right answer. Where did I make a mistake?
    You need to use the product rule or quotient.

    \frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    You need to use the product rule or quotient.

    \frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}

    I got it.
    Thank you!
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