Expert: Alon Mandes - 11/1/2009

Question
Investigate y=lnx/1-lnx.Graphic needed too

Answer
The investigation process of the graph y=Ln(x)/[1-Ln(x)] will be as follow :

Step #1
Finding points of intersects with the x %26 y axis :
Intersect with x-axis : we solve the equation y=0 : Ln(x)/[1-Ln(x)]=0 --> Ln(x)=0 --> x=1 .
Intersect with y-axis : None . There exists only vertical asymptotes

Step #2
Finding Asymptotes :
1st Horizontal asymptotes :
y=Lim  Ln(x)/[1-Ln(x)]=Lim  1/[1/Ln(x) -1]= 1/[1/Ln(∞) -1]=1/[0-1]=-1
                         x->∞                    x->∞
2nd Horizontal asymptotes :
y=Lim  Ln(x)/[1-Ln(x)]=Lim  1/[1/Ln(x) -1]= 1/[1/Ln(0) -1]=1/[1/-∞ -1]=1/[0-1]=-1
  x->0                     x->0
Vertical Asymptotes : solving the equation 1-Ln(x)=0 : Ln(x)=1 --> x=e=2.71

Step #3
Find domain %26 range of the function :
Domain : x≥0
Range  : y≤1

Step #4
Finding Critical points (extremum) :
We solve the equation y'=0 : y'={ [1/x]+[Ln(x)/x] }/{ 1-Ln(x) }²
y'=0 --> { [1/x]+[Ln(x)/x] }/{ 1-Ln(x) }² = 0 --> { [1/x]+[Ln(x)/x] } = 0  -> Ln(x)+1=0 .
No positive value of x can make the equation stand true. Therefore, there is no critical
points. This means that the function is monotonic %26 is always rising .

Step #5
Finding intervals of positivity %26 negativity :
The function is positive in the interval : Ln(x)/[1-Ln(x)]>0 This is ture when 1<x<e .
The function is negative in the interval : Ln(x)/[1-Ln(x)]<0 This is ture when 0<x<1 or x>e .

Step #6
Graphing the function :
The function starts from x=0 with value of y=-1 , %26 start to rise slowly towards infinity.
at x=1 the function intersect with the x-axis, %26 then the function continue rising %26 reaches
the infinity at x=e. At x>e the function rises again from -infinity %26 start to rise slowly towards the value y=-1 .


Alon.