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

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.

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

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

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

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


f'(x) = \frac {ln(x)(2 - ln(x))} {x^2}


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:


\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.

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

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