Graphing question (1 Viewer)

[johNNyboy772]

can someone graph for me lnx / x ?

How can we find out whehter at the origin the graph approaches origin or neg-infinite? without using the calculator to sub a point in

 

[scroates]

as x approaches zero lnx approaches negative infinity and x approaches zero hence

 

[untouchablecuz]

graph y₁=Ln(x) and y₂=x:

- x>0 since Ln(x) does not exist for x≤0
- from the graph, y₂>y₁ for all x. hence, x>Ln(x) and thus Ln(x)/x<1
- intercepts the x axis at x=1
- for 0< x<1, Ln(x) is -tve and x is +tve. hence Ln(x) is -tve in this domain
- for x>1 Ln(x) is +tve and x is +tve. hence Ln(x) is +tve in this domain
- as x-->0, Ln(x)-->-infinity and x-->0 and thus Ln(x)/x-->-infinity

also, y' = [x(1/x)-1*Ln(x)]/(x²)=[1-Ln(x)]/x²

stationary point at x=e, y=1/e. it is a maximum (deduced from above info)

 

[Lukybear]

use the calculataor. Except you dont say use a calculator, but put

as x>>> inf, y>>> 0 etc...

Also, i think that is division? Draw both graphs and calculate.

 

[Lukybear]

Untouchable is that how you would answer in exam?