find all extremum points

**cgiulz** · November 2nd 2009, 06:23 PM

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

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

**VonNemo19** · November 2nd 2009, 06:25 PM

Originally Posted by cgiulz

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

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

What are you having trouble with here? What have you tried? Is it taking the derivative? Choosing test values? etc...?

**cgiulz** · November 2nd 2009, 06:45 PM

Try them out and you'll see. For $(a)$ the criticals are 0,1 and -1, and as $x \rightarrow \infty, f(x) \rightarrow 0,$ so might this be a minimum? Also, +-1 seem to be a single max? Moreover, the first and second derivative test fail..etc

**cgiulz** · November 2nd 2009, 06:50 PM

I think I got $(a)$ -- 0 is a max and +-1 are neither.

**VonNemo19** · November 2nd 2009, 06:54 PM

Originally Posted by cgiulz

Try them out and you'll see. For $(a)$ the criticals are 0,1 and -1, and as $x \rightarrow \infty, f(x) \rightarrow 0,$ so might this be a minimum? Also, +-1 seem to be a single max? Moreover, the first and second derivative test fail..etc

a) $f'(x)=\frac{2x(1+x^4)-4x^3(x^2)}{(1+x^4)^2}=0\Rightarrow{x(1-x)(1+x)(1+x^2)}=0\Rightarrow$ critical numbers at $x=1,-1,0$

So, evaluate f prime within these intervals and determine its sign.

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