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
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
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?