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saywhat234 · Apr 27, 2010, 04:32 AM

f(x)=ln²x/x
Find local maxima & minima and then find all inflection points. Find asymptotes(vertical or horizontal) after stating the domain of the given function

ebaines · Apr 27, 2010, 10:10 AM

I assume you know that the local minima and maxima can be determined by solving the equation f'(x) =0, and that the inflection points are at f''(x) = 0. I suggest you try to solve this and then post back with your results - we'll be happy to check it for you.

saywhat235 · Apr 27, 2010, 05:20 PM

OK I got the local minimum at(1,0) and no local max.. the inflection pt at(2.61,0.35), the domain of the function is f(x)>0.. there is a vertical and horizontal assypmtote but I'm having trouble finding the limits.. can u please tell me if my answers are correct and if not tell me what the right answers are and the shape of the graph.. I have been working on this problem for an entire week and still didn't come up with the right answers.. thanks a lot

saywhat235 · Apr 27, 2010, 05:20 PM

OK I got the local minimum at(1,0) and no local max.. the inflection pt at(2.61,0.35), the domain of the function is f(x)>0.. there is a vertical and horizontal assypmtote but I'm having trouble finding the limits.. can u please tell me if my answers are correct and if not tell me what the right answers are and the shape of the graph.. I have been working on this problem for an entire week and still didn't come up with the right answers.. thanks a lot

saywhat234 · Apr 28, 2010, 03:03 AM

Please can you help me with this as soon as possible? I'm trying so hard to find the correct answers... thank you

ebaines · Apr 28, 2010, 06:07 AM

In addition to the minimum that you found, there is also a maximum. From the dereivative:

$<br /> f'(x) = \frac {ln(x)(2 - ln(x))} {x^2}<br />$

you can see that this is equal to zero at ln(x) = 0 and ln(x) = 2.

The domain is x>0 (because ln(x) is limited to only positjve values of x), and the range is y>=0 (since ln^2(x) can't be negative). There are two asymtotes - one is pretty obviously the vertical at x = 0. To find the other, consider what happens in the limit as x goes to infinity. Use l'Hospital's rule:

$<br /> \displaystyle \lim _ {x \to \infty} \frac {ln^2x} {x} = \frac { \lim _ {x \to \infty} 2 ln(x) \frac 1 x } {\lim _ {x \to \infty} 1 } = \frac {\lim _ {x \to \infty}2 ln(x)} x = 0.<br />$

saywhat234 · Apr 28, 2010, 06:18 AM

Thanks a lot... I managed to figure out the whole problem after a lot of hard work.. I just want to make sure that my graph is correct.. how does the graph look like.. thanks again

ebaines · Apr 28, 2010, 06:48 AM

Do you have a graphing calculator? Or you can do what I do - set up a spreadsheet.