f(x) = , g(x) = x² + 1

Now,

Domain of f = [ – 3, ∞], domain of g = (– ∞, ∞)

Range of f = [0, ∞), range of g = [1, ∞)

Then, range of f ⊂ Domain of g and range of g ⊂ Domain of f

Hence, fog and gof exists

Now,

fog(x) = f(g(x)) = f(x² + 1)

⇒ fog(x) =

Again,

gof(x) = g(f(x)) = g(

⇒ gof(x) =

⇒ gof(x) = x + 4