f(x) = x + 1 and g(x) = e^x

Range of f = R ⊂ Domain of g= R ⇒ gof exist

Range of g = (0, ∞) = Domain of f = R ⇒ fog exist

Now,

gof(x) = g(f(x)) = g(x + 1) = e^{x + 1}

And

fog (x) = f(g(x))= f(e^x) = e^x + 1

Hence, fog(x) = e^{x +} 1 and gof(x) = e^x + 1