Calculus
posted by Jessie on .
Let h(x) = 4x^2 + 2 x + 2.
Let g(x) = 3√{x} + 1 (This is the third square root plus 1)
(a) g(h(x)) =
(b) h(g(x)) =
(c) h(h(x)) =
(d) g(x) + 1 =
(e) g(x+1) =

There's no such thing as "third square root". There's "square root", "cube root", "fourth root", etc.
g(h) = ∛h+1 = ∛(4x^2+2x+2)+1
h(g) = 4g^2 + 2g + 2
= 4(∛x+1)^2 + 2(∛x+1) + 2 = 4∛x^2 + 10∛x + 8
h(h) = 4h^2 + 2h + 2
= 4(4x^2 + 2 x + 2)^2 + 2(4x^2 + 2 x + 2) + 2
= 64x^4 + 64x^3 + 88x^2 + 36x + 22
g(x)+1 = (∛x+1)+1 = ∛x + 2
g(x+1) = ∛(x+1) + 1