Помогите пожалуйста решить. Помогите пожалуйста исследовать фурнкцию y=lnx/x. Заранее спасибо.

First, find intercepts:
y-intercept ---> set x = 0, undefined
x-intercept ---> set y= 0, ln(x)/x = 0 ---> ln(x) = 0 ---> e^0 = x
x = 1 ----> x-intercept is (1, 0)

Asymptotes:
lim (x --> inf.) ln(x)/x ----> x goes to infinity much faster than ln(x), so the limit goes to zero..
so y = 0 is an asymptotes...

Observe from above, when finding the y-intercept, the function cannot be evaluated at x = 0, so that too is an asymptote...

Now, derivatives:

y = ln(x)/x
y' = [1 - ln(x)] / x^2
y" = [-3 + 2ln(x)] / x^3

Set y' = 0 = [1 - ln(x)] / x^2
[1 - ln(x)] = 0, ln(x) = 1 and x = e

Pluggin into the second derivative:
y"(e) = [-3 + 2ln(e)] / (e)^3 = -e^-3
Since y" < 0, y has a max at x = e
So, y(e) = 1/e. The max is at point (e, 1/e)

For point of inflection, set y" = 0
y" = [-3 + 2ln(x)] / x^3 = 0
[-3 + 2ln(x)] = 0
ln(x) = (3/2)
x = e^(3/2)
y(e^(3/2)) = (3/2)*e^(-3/2)
Point of inflection is at [ e^(3/2) , (3/2)*e^(-3/2) ]

For intervals of increasing/decreasing,
The function is decreasing when y' <0

y' = [1 - ln(x)] / x^2,
y' < 0 if ln(x) > 1, x > e

The function is increasing when y' > 0
y' > 0 if ln(x) < 1, x < e

For concavity,
The function is concave up when y" >0
y" > 0 if [-3 + 2ln(x)] > 0, ln(x) > (3/2)
x > e^(3/2)

The function is concave down when y" < 0
y" < 0 if [-3 + 2ln(x)] < 0, ln(x) < (3/2)
x < e^(3/2)