We simply substitute:
g ( x ) = ( f ∘ f ) ( x ) = f ( f ( x ) ) = f ( x − 1 ) g ( x ) = x − 1 − 1. {\displaystyle {\begin{aligned}g(x)&=(f\circ f)(x)\\&=f(f(x))\\&=f({\sqrt {x}}-1)\\g(x)&={\sqrt {{\sqrt {x}}-1}}-1.\end{aligned}}}
