# find all extremum points

• Nov 2nd 2009, 06:23 PM
cgiulz
find 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 PM
VonNemo19
Quote:

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

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

What are you having trouble with here? What have you tried? Is it taking the derivative? Choosing test values? etc...?
• Nov 2nd 2009, 06:45 PM
cgiulz
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 PM
cgiulz
I think I got $\displaystyle (a)$ -- 0 is a max and +-1 are neither.
• Nov 2nd 2009, 06:54 PM
VonNemo19
Quote:

Originally Posted by cgiulz
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

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

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