4. (f∘g)(x) = f(g(x)) = [g(x)]^3 + [g(x)]^2 = [g(x)]^2*([g(x)]+1) = (4x-9)^2*(4x-8) = 4(x-2)(16x^2 - 72x + 81) = 4(16x^3 + 40x^2 + 225x -162). but g∘f(x) = g(f(x)) = 4f(x) - 9 = 4x^3 + 4x^2 - 9. So the operation is NOT commutative.
5. Suppose "one-to-one" means "injective", U(t1) = U(t2) only if t1 = t2. So let (g∘f)(x1) = (g∘f)(x2), g(f(x1)) = g(f(x2)). g is one-to one implies that f(x1) = f(x2). Now f is one-to-one gives that x1 = x2, which we ought to prove.
(if "one-to-one" means "bijective" than we use inverses for f and g, I'll write if you want)
Apr 27th, 2015
Did you know? You can earn $20 for every friend you invite to Studypool!