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!
