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Summer Clouds moderators

22/07/2017 at 12:01
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Given \(H=\dfrac{1}{x^2-2x+5}\). Find the value of x such that H reaches the maximum value.


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    Kayasari Ryuunosuke Coordinator 22/07/2017 at 12:10

    We have :

    \(x^2-2x+5=x^2-2x+1+4\)

                          \(=\left(x-1\right)^2+4\ge4\)

    \(\Rightarrow H=\dfrac{1}{x^2-2x+5}\le\dfrac{1}{4}\)

    \(\Rightarrow Max_H=\dfrac{1}{4}\)

    \(\Leftrightarrow\left(x-1\right)^2=0\)

    \(\Leftrightarrow x=1\)

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    ¤« 03/04/2018 at 13:33

    We have :

    x2−2x+5=x2−2x+1+4

                          =(x−1)2+4≥4

    ⇒H=1x2−2x+5≤14

    ⇒MaxH=14

    ⇔(x−1)2=0

    ⇔x=1


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