Is the discontinuity of f(f(f(x))) at x=0 and x=1 non-removable?

[phymatter]

if f(x)=1/(1-x) , then the points of discontinuity of f(f(f(x))) are x=0 and x=1 ,
b'cos f(f(x)) is (x-1)/x and f(f(f(x))) is x , but my book (I.A. Maron S.V.C) says that both x=0 and x=1 are removable discontinuities , but i don't think so! pl.confirm me!

 

[Office_Shredder]

When you calculated f(f(f(x))) if you canceled anything, then you were removing a removable singularity

 

[JonF]

Neither are actually discontinuities since they are removed from the domain of the function. But that’s largely semantic. The idea they are getting at I think is best understood with an example.

If f(x) = (x-1)(x+1)/(x-1) and g(x) = x+1 with standard treatment of the domain are they same? The answer is no, because g(1) = 2 and f(1) is undefined.

You have to be careful when you do algebraic manipulations to functions of how you alter the domain. so when you got f(f(x)) = (x-1)/x, x=0 is removed from the domain at that point.

 

[dreit]

So this discontinuity would be non-removable?