Given that y=lnx/x for x>0, find the set of values of x for which y is an increasing function of x
Using Calculus, you don't even have to think to solve this problem. Simply take the derivative and find the critical point to get the extrema, which happens to be 1 and e (over the given interval).
This can be solved without calculus by simply thinking about the function. The function is Lnx/x, so the denominator is dominant. Compare this to the function x/x which is always 1.
We can know that the starting value (minimum) is going to be 1, since the Ln1 = 0. Less than 1 will get you negative numbers.
The maximum will occur when the numerator becomes greater than 1, when Lnx > 1. This is because once you get over one in the numerator, the ratio will switch over and your divisions will get smaller and smaller. So what value for x will get you Lnx = 1? Lne of course.
So the set is from 1 to e the function is increasing, and then greater than e the function is decreasing. There are no positive values for the function when x < 1.
Patrick
EDIT: I originally misread the question. I thought the question wanted where the function values are > 0, rather than x > 0. The answer can be changed because the function is still increasing when x < 1. The function has a vertical asymptote at zero, since that would be undefined. So the answer would be (0, e].