If $f(x)=\frac{1}{1-x}$ and $f^n=$ fofof .... of, then the points of discontinuity of $f^{3 n}(x)$ is/are

x = 0, 1

Clearly, x = 1 is a point of discontinuity of f(x).

Now, for x ≠ 1, we have

fof(x) = f(f(x))

⇒ fof(x) = $f\left(\frac{1}{1-x}\right)=\frac{1}{1-\frac{1}{1-x}}$

⇒ fof(x) = $\frac{x-1}{x}$, which is discontinuous at x = 0.

If x ≠ 0, 1, then

(fofof)(x) = f(fof(x)) = $f\left(\frac{x-1}{x}\right)$

⇒ fof(x) = $\frac{1}{1-\frac{x-1}{x}}=x$, which is everywhere continuous.

Hence, $f^{3 n}(x)=\left\{\begin{array}{l}\left(\begin{array}{l}((fofof) o (fofof) ... o (fofof) \\ ~~~~~~~~\quad n-\text { times }\end{array}\right)(x) \\ x, ~~\text{which is everywhere continuous}\end{array}\right.$

So, the only points of discontinuity of $f^{3 n}(x)$ are x = 0 and x = 1.