# Math Help - Global maximum and minimum

1. ## 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?

2. ## Re: Global maximum and minimum

You are on the right track, I'd sketch the function first, maybe that will give you some hints.

3. ## 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 $\min{\left\{\ g(x)\right\}}=\frac{1}{\max{\left\{ f(x)\right \}}}$

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

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