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Math Help - Finding max mins point of inflection.

  1. #1
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    Finding max mins point of inflection.

       f(x)= 2x^2 + \frac {8}{x^2}

      f'(x)= 4x - \frac{16}{x^3}

      f''(x) = 4+\frac{48}{x^4}

    So taking f''(x)

     \frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}

    \frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4

     4x^4+48 = 0

     48=-4x^4  => -12 = x^4 => -1.86 = x

    so i have a point of inflection at -1.86?

    then evaluate f(-1.86).

    2\cdot -1.86^2 + \frac{8}{-1.86^2} = -9.23

    So this is a local maximum?

    somthing seems wrong , can anyone point out what it is?
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  2. #2
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    Quote Originally Posted by el123 View Post
       f(x)= 2x^2 + \frac {8}{x^2}

      f'(x)= 4x - \frac{16}{x^3}

      f''(x) = 4+\frac{48}{x^4}

    So taking f''(x)

     \frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}

    \frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4

     4x^4+48 = 0

    \textcolor{red}{4x^4 + 48 = 0} has no real solutions.

    f''(x) > 0 for all x in the domain of f, so f is concave up everywhere in its domain, i.e. no inflection points.
    ...
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  3. #3
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    is that because  \sqrt[4]{-12} doesnt exist?

    Are you able to tell me or direct me to somewhere where i may learn the process of finding max/mins , i have been trying to find somwhere but i can't. My textbook is worthless.Im getting really frustrated trying to work these out i just don't understand them.
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  4. #4
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    is that because  \sqrt[4]{-12} doesnt exist?

    Are you able to tell me or direct me to somewhere where i may learn the process of finding max/mins , i have been trying to find somwhere but i can't. My textbook is worthless.Im getting really frustrated trying to work these out i just don't understand them.
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