A composition of functions roughly means that one function is being applied to the result of the other.
Let's start with #f @ g#. This composition states that first, #g(x)# is being computed and later, this result is being taken as an "input" to the function #f#:
#(f @ g)(x) = f(g(x)) = f( 4/5 x)#
This means that instead of #x#, the function #f# now takes # (4/5 x)# as input. So, you need to plug #4/5 x# for every occurance of #x#:
#(f @ g)(x) = f(g(x)) = f( 4/5 x) = 5/ (4/5 x - 6) = 5 / ((4x - 30)/6) = 30 / (4x - 30)#
Now, let's build the other composition, # g @ f #. This time, first #f(x)# is being computed and later, the result needs to be taken as "input" for #g#:
# (g @ f) (x) = g(f(x)) = g(5/(x-6))#
... plug # 5/(x-6)# for every occurance of #x# in #g(x)#...
# (g @ f) (x) = g(f(x)) = g(5/(x-6)) = 4/5 * 5/(x-6) = 4 / (x-6)#
Hope that this helped!
