Question

Given the function $f\left(x\right)=\frac{1}{x-1}.$ Find the number of points of discontinuity of the composite function $f\left[f\left\{f\left(x\right)\right\}\right].$

Answer 1

The point $x=1$ is clearly a point of discontinuity of the function $y=f\left(x\right)=\frac{1}{1-x}.$
If $x\ne 1,$ then $v\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{1}{1-x}\right)=\frac{1}{1-\left[1/(1-x)\right]}=\frac{x-1}{x}$
Hence, the point $x=0$ is a discontinuity of the function $v$.
If $x\ne 0,x\ne 1,thenw\left(x\right)=f\left[f\left\{f\left(x\right)\right\}\right]=f\left[f\left(\frac{1}{1-x}\right)\right]=f\left(\frac{x-1}{x}\right)=\frac{1}{1-(x-1)/x}=x.$
Hence w is clearly continuous everywhere.
Thus,the points of discontinuity of the composite function $f\left[f\left\{f\left(x\right)\right\}\right]arex=0.x=1$ and the composite function $f\left\{f\left(x\right)\right\}$ has a discontinuity at $x=1$ only.