f(x)=(lnx)^2, x>0

On this particular problem i don't even know how to start so if you would assist me in starting this problem that would be awesome. just help me start and i'm sure i could probably figure out the rest

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- Sep 24th 2008, 02:46 PMplstevensfind the value of any relative extrema.
f(x)=(lnx)^2, x>0

On this particular problem i don't even know how to start so if you would assist me in starting this problem that would be awesome. just help me start and i'm sure i could probably figure out the rest - Sep 24th 2008, 03:19 PMskeeter
start by finding f'(x) and looking for critical vales ...

$\displaystyle f'(x) = \frac{2\ln{x}}{x}$

$\displaystyle 2\ln{x} = 0$

$\displaystyle x = 1$

for $\displaystyle 0 < x < 1$ , $\displaystyle f'(x) < 0$ ... f(x) is decreasing

for $\displaystyle x > 1$, $\displaystyle f'(x) > 0$ ... f(x) is increasing

so ... what sort of extrema at x = 1?