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Math Help - Global maximum and minimum

  1. #1
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    Global maximum and minimum

    I am having trouble with finding the global maximum and minimum values of this function:

    \frac {1} {(x-8)^2+6}

    Am I supposed to be finding the derivative of it and then setting it to zero or I'm I thinking the completely wrong thing here?
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  2. #2
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    Re: Global maximum and minimum

    You are on the right track, I'd sketch the function first, maybe that will give you some hints.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Global maximum and minimum

    Find global minimum and maximum of \frac{1}{(x-8)^2+6}
    Let f(x)=(x-8)^2+6 and g(x)=\frac{1}{(x-8)^2+6}

    You must understand that:

    \max{\left\{\ g(x)\right\}}=\frac{1}{\min{\left\{ f(x)\right \}}} and \min{\left\{\ g(x)\right\}}=\frac{1}{\max{\left\{ f(x)\right \}}}

    ________________________________________________

    \min{\left\{ (x-8)^2+6\right\}}=6 \ \text{{\color{red}at}} \ x=8...<------Why? Find out by using your knowledge of maxima and minima.

    \max{\left\{ \frac{1}{(x-8)^2+6}\right\}}=\frac{1}{6} \ \text{{\color{red}at}} \ x=8
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