SOLUTION: What is the interval where f(x)=lnx/x (x>0) decreases?

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Question 1089124: What is the interval where f(x)=lnx/x (x>0) decreases?
Found 2 solutions by Theo, rothauserc:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
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.

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 http://www.derivative-calculator.net/

Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
f(x) = lnx/x
:
f'(x) = (1 - ln(x)) / x^2
:
set f'(x) = 0 and solve for x
:
(1 - ln(x)) / x^2 = 0
ln(x) = 1
x = e
:
at x = e, f(x) = 1/e, this is a global maximum
:
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f(x) decreases on (e, +infinity)
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Here is a graph of f(x)
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