Question 1089124
the derivative of ln(x) / x is equal to (ln(x) - 1) / x^2


set this equal to 0 and you'll find a max/min point on the equation of y = ln(x) / x.


start with (ln(x) - 1) / x^2 = 0


multiply both sides of the equation by x^2 to get ln(x) - 1 = 0


add 1 to both sides of the equation to get ln(x) = 1


this is true if and only if e^1 = x


solve for x to get x = e which is equal to 2.718281828


that's a max/min point on the graph.


looking at the graph, you see that it's a max point.


this means the value of ln(x) / x will be increasing to the left of this max point and decreasing to the right of this max point.


the interval where ln(x)/x is decreasing is therefore from x = 2.718281828..... to positive infinity.


in interval notation, i believe this will be [2.718281828..., positive infinity)


the following graph shows this to be true.


the graph rounds the solution to 3 decimal places.


<img src = "http://theo.x10hosting.com/2017/072904.jpg" alt="$$$" </>


as x gets larger, the function ln(x) / x will approach 0 but will never reach 0.


note that i didn't figure out what the derivative was.


i probably could if i remembered how to do it.


since i didn't, i cheated and used an online derivative calculator.


it works very well and can be quite useful for you to see if you calculated the derivative correctly.


that calculator can be found at <a href = "http://www.derivative-calculator.net/" target = "_blank">http://www.derivative-calculator.net/</a>