Let and . We need to find f(x) such that f(g(x)) = h(x).

If g(x) were a bijection, then there would exist a unique such that for all . Then we could represent any y as and get . But in this case, g(x) is neither surjective (if the codomain is ) nor injective. More precisely, g(x) is not surjective because g(x) ≤ 3 for all x, so we have no information about f(y) when y is not in the image of g, i.e., when y > 3. And g(x) is not injective because there are two possible values of for each y < 3. Unfortunately, h(x) returns different results for these two values. This means we can't assign f(y) that would work for both possible values of . For example, g(-1) = g(-3) = 0, so f(0) has to equal both h(-1) = -2 and h(-3) = 18. If we set f(0) = -2, then f(g(x)) = h(x) is satisfied for x = -1 but not for x = -3, and if we set f(0) = 18, then f(g(x)) = h(x) works for x = -3, but not for x = -1.

The most we can do is to define, say, for y ≤ 3 using the right branch of g(x), i.e., for x ≥ -2. Then f(g(x)) = h(x) will hold for x ≥ -2.