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