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[conman]

Sketch y = lnx/x
I found the first derivative = (1 - lnx)/x^2
Domain x>0 which is x= 0 is vertical tangent.
y' = 0 result x = e.
lim (x - to - infinity and + infinity) = 0 horizontal tangent.
Check the sign y'= 1 - lnx

It made me confused should x is positive or negative???

 

[SoulSearcher]

y = 0, lnx / x = 0
i.e. ln x = 0
x = 1

In this case, and for most logarithm graphs, x must be positive.

And the graph 1 - ln x is a different graph.

EDIT: sorry, my mistake

 

[conman]

y = 0, lnx / x = 0
i.e. ln x = 0, which isn't possible.

In this case, and for most logarithm graphs, x must be positive.

And the graph 1 - ln x is a different graph.

There is a mistake in picture; I mean y'=0 not y=0
I used the graph 1 - ln x to find the sign of the main graph (when it goes up and down and to find critical point).

 

[SoulSearcher]

SInce x² is always positive, check sign of 1 - ln x for x > 0. x cannot exist < than 0.

0 < x < e, 1 - ln x > 0, therefore it has positive gradient.
x = e, 1 - ln x = 0, therefore it is a turning point.
x > e, 1 - ln x > 0, therefore it has negative gradient.

Therefore the point (e, 1/e²) is a maximum turning point.

Forgetting the basics?

What is ln 1?

Yeah, I forgot that

 

[conman]

SInce x² is always positive, check sign of 1 - ln x for x > 0. x cannot exist < than 0.

0 < x < e, 1 - ln x > 0, therefore it has positive gradient.
x = e, 1 - ln x = 0, therefore it is a turning point.
x > e, 1 - ln x > 0, therefore it has negative gradient.

Therefore the point (e, 1/e²) is a maximum turning point.
Yeah, I forgot that

Yeah I forgot that x >0
And my graph is wrong too; it has to be reflect around the x -axis not y-axis and moving up 1 unit. I got it now. Thanks