$\text{ If }f\left(x\right)\text{ is continuous and increasing function such that domain of }g\left(x\right)=\sqrt{f\left(x\right)-x}\text{ be }R\text{ and }h\left(x\right)=\frac{1}{1-x},\text{ then }$

$\text{ the domain of }\varphi \left(x\right)=\sqrt{f\left(f\right(f\left(x\right)\left)\right)-h\left(h\right(h\left(x\right)\left)\right)}\text{ is }$

$$\begin{gathered} h\left(x\right) =\frac{1}{1-x} ,x\ne 1 \\ \Rightarrow h\left(h\left(x\right)\right)= \frac{1}{1-{\displaystyle \frac{1}{1-x}}}= \frac{x-1}{x}, x\ne 0,1 \end{gathered}$$

$\Rightarrow h\left(h\left(h\left(x\right)\right)\right)=\frac{1}{1-\left({\displaystyle \frac{x-1}{x}}\right)}=x , x\ne 0,1$

$$\begin{gathered} \Rightarrow g\left(x\right) =\sqrt{f\left(x\right)-x}\ge 0 ,for all x\in R \\ \Rightarrow f\left(x\right)\ge x \Rightarrow f\left(f\left(x\right)\right)\ge f\left(x\right)\ge x \left(\sin ce f\left(x \right)is increa\sin g function\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow similarly f\left(f\left(f\left(x\right)\right)\right)\ge x \\ \Rightarrow f\left(f\left(f\left(x\right)\right)\right)-x \ge 0 , for all x\in R-\left\{0,1\right\} \\ \Rightarrow \phi \left(x \right) =\sqrt{f\left(f\left(\left(f\left(x\right)\right)\right)\right)-x } is defined for all x\in R-\left\{0,1\right\} \end{gathered}$$