finding f(g(x)) and g(f(x))

Question

Can somebody help me find the f(g(x)) and g(f(x)) for all of these problems? I need both for each.

1. 1.    f = 3x and g= 1/3 x
2. 1.    f = 4x -2 and g= x+2/4
3. f = 5x - 1 and g=  x/5 + 1
4. 1.    f =  3/4 x - 6    and g= 4/3 x+ 8 
5. f = 2x+18/3 and g = 3/2 x - 9   
6.       1.  f = x² + 5 and g = √x -5

I made things like x and -5 BOLD because they don't go inside the fraction or square root.

Answer 1

1. f(g(x)) = f(x/3) = 3(x/3) = x

g(f(x) = g(3x) = 3x/3 = x

they're inverse functions

2. This is almost an inverse function, which suggests maybe it was copied incorrectly or a typo

f(g(x)) = f(x+1/2) = 4(x+1/2) - 2 = 4x +2

g(f(x)) = g(4x-2) = 4x-2 + 1/2 = 4x -3/2

IF g(x) had been = x/4 + 2/4, they would have been inverse functions and f(g(x))=g(f(x)) = x

3.this is an inverse function: f(g(x))= g(f(x)) = x

4.another inverse function, both = x again

5.f(g(x)) = f(3x/2-9) = 3x-18 + 6 = 3x-12

g(f(x)) = g(2x+6)= 3(2x+6)/2 -9 = 3x

6 another possible typo? It would be an inverse relation if that -5 were under the radical

but as is f(g(x)) = f(sqrx -5) = (sqrx-5)^2 + 5 = x -10sqrx +25 + 5 = x-10sqrx +30

g(f(x)) = g(x^2 +5) = sqr(x^2+5) -5

Just slightly different problem on a couple of these, and everyone of them would be f(g(x)) = g(f(x)) = x

another question is why is this listed under the category "trigonometry"?