$\displaystyle (a) \frac{x^2}{1 + x^4}$ over $\displaystyle (-\infty,\infty).$

$\displaystyle (b) \frac{\ln(x)}{x}$ over $\displaystyle [1,\infty).$

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- Nov 2nd 2009, 06:23 PMcgiulzfind all extremum points
$\displaystyle (a) \frac{x^2}{1 + x^4}$ over $\displaystyle (-\infty,\infty).$

$\displaystyle (b) \frac{\ln(x)}{x}$ over $\displaystyle [1,\infty).$ - Nov 2nd 2009, 06:25 PMVonNemo19
- Nov 2nd 2009, 06:45 PMcgiulz
Try them out and you'll see. For $\displaystyle (a)$ the criticals are 0,1 and -1, and as $\displaystyle x \rightarrow \infty, f(x) \rightarrow 0,$ so might this be a minimum? Also, +-1 seem to be a single max? Moreover, the first and second derivative test fail..etc

- Nov 2nd 2009, 06:50 PMcgiulz
I think I got $\displaystyle (a)$ -- 0 is a max and +-1 are neither.

- Nov 2nd 2009, 06:54 PMVonNemo19