#f(x)# is algebraic function #sqrt(1–x) = (1–x)^(1/2)# while #g(x)# is transcendental function (logarithmic), since algebraic operations transform algebraic expressions into algebraic expressions and transcendental expressions into transcendental expressions but not algebraic expressions into transcendental expressions.

For a composite function #f\circg# aka #f(g(x))# to exist, the range of #g#, #R_g# (i.e. set of values of #y# when #y = g(x)#) must be smaller than or equal to (i.e. a subset) the domain of #f#, #D_f# (i.e. the #x# values that are 'plugged' into the equation #y = f(x)#). That is, #R_g \sube D_f#.

Since the question does not state the domain or range of #f# (i.e. #D_f#) or #g# (i.e. #D_g#), I will assume that #D_f = [1, \infty)#, that is, all real numbers greater than or equal to 1 and #D_g = (–\infty, 0) \cap (2, \infty)#, that is, all real numbers small than 0 or greater than 2 (to allow whatever is in the #ln# function to be positive). Thus, plugging the sets of values of #x# into #f# and #g# produces #R_f = RR_0^+# (that is, all non-negative real numbers) and #R_g = RR# (all real numbers).

In this case, #R_f = RR_0^+ \sube (–\infty, 0) \cap (2, \infty) = D_g# is not true, since #RR_0^+# contains numbers that are not in #(–\infty, 0) \cap (2, \infty)# (e.g. #[0, 2]#), thus #g\circf# (i.e. #g(f(x))#) does not exist. #R_g = RR \sube [1,∞) = D_f# also doesn't hold as it includes elements absent in #D_f# (i.e. #(–\infty, 1)#), thus #f\circg# aka #f(g(x))# also does not exist.