#(f@g)(x)# is equivalent to #f (g(x))#. We solve this problem just as we solve #f(x)#. But since it asks us to find out #f(g(x))#, in #f(x)#, each time we encounter x, we replace it with #g(x)#.
In the above problem, #f(x) = x+3#.
Therefore, #f(g(x))= g(x) +3#.
#=> (f@g)(x)= 2x - 7 + 3#
#=> (f@g)(x)= 2x - 4#
Basically, write the #g(x)# equation where you see the x in the #f(x)# equation.
#f@g(x)= (g(x))+3# Replace #g(x)# with the equation
#f@g(x)= (2x-7)+3#
#f@g(x)= 2x-7+3# we just took away the parentheses
#f@g(x)= 2x-4# Because the #-7+3=4#
This is it
#g@f(x)# would be the other way around
#g@f(x)= 2(x+3)-7#
now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.
#g@f(x)= 2x+6-7#
Next, #+6-7 = -1#
#g@f(x) = 2x-1#
Its a lt easier than you think!
