If f(x) = x + 1 and g(x) = x + 3 then f ∘ f ∘ f ∘ f is :

Each application of f adds 1, so $f\circ f\circ f\circ f(x) = x + 4$. Since $g(x) + 1 = (x + 3) + 1 = x + 4$, the composition equals g + 1. g alone gives only x + 3, and the four-fold composition of g would give x + 12.