# Thread: help me solve this y=lnx/x x>0

1. ## help me solve this y=lnx/x x>0

Given that y=lnx/x for x>0, find the set of values of x for which y is an increasing function of x

2. Originally Posted by Joelyks
Given that y=lnx/x for x>0, find the set of values of x for which y is an increasing function of x
$y=\frac{\ln(x)}{x}$

$y'=\frac{x\frac{1}{x}-\ln(x)}{x^2}=\frac{1-\ln(x)}{x^2}$

A function is increasing when $f'(x) > 0$

Just solve from here.

3. 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].

4. oh thanks
So is the solution : lnx<1
x<e^1
x<2.72 ( 3s.f) ?

5. Yep, just don't forget the lower bound of zero (exclusive).