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Math Help - Local Maximum

  1. #1
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    Local Maximum

    Prove that the function f(x)=x^2+(a/x) cannot have a local maximum for any value of a.

    Any help would be greatly appreciated thanks.
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  2. #2
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    Hello, Sophia27!

    Prove that the function f(x)\:=\:x^2+\frac{a}{x} cannot have a local maximum for any value of a.
    We have: . f(x)\:=\:x^2 + ax^{-1}

    Then: . 2x -ax^{-2} \:=\:0

    Multiply by x^2\!:\;\;2x^3 - a \:=\:0\quad\Rightarrow\quad x^3 \:=\:\tfrac{a}{2} \quad\Rightarrow\quad x \:=\:\sqrt[3]{\tfrac{a}{2}} .
    (critical value)


    Second derivative: . f''(x) \;=\;2 + 2ax^{-3} \:=\:2 + \frac{2a}{x^3}

    Then: . f''\!\left(\sqrt[3]{\tfrac{a}{2}}\right) \;=\;2 + \frac{2a}{\left(\sqrt[3]{\frac{a}{2}}\right)^3} \;=\;2 + \frac{2a}{\frac{a}{2}} \;=\;2 + 4 \:=\:{\color{red}+}6

    At our critical value, the second derivative is positive; the graph is concave up, \cup

    We have one critical value and it is a local minimum.


    Therefore, the function has no local maximum value.

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