39As per the definition of [f(x)],
1 ≤ f(x) < 2
=> 1 ≤ arccos(arcsin(arctan(x))) < 2
=> cos1 ≤ arcsin(arctan(x)) < cos2
=> sin(cos1) ≤ arctan(x) < sin(cos2)
=> tan(sin(cos1)) ≤ x < tan(sin(cos2))
Therefore x → [ tan(sin(cos1)) , tan(sin(cos2)) )