Let f:R→R, g:R→R and h:R→R be differentiable functions such that f(x) = x³ + 3x + 2, g(f(x)) = x, and h(g(g(x))) = x for all x ∈ R. Then g'(2) = 1/15, h'(1) = 666, h(0) = 16, h(g(3)) = 36.

Here, f(x) = x³ + 3x + 2 and, g(f(x)) = x and, f'(x) = 3x² + 3
g(f(x)) = x
Differentiating both sides, we get
g'(f(x))f'(x) = 1 ⇒ g'(f(x)) = 1/f'(x)
Let f(x) = 2 at x = 0
∴ g'(f(0)) = g'(2) = 1/f'(0) ⇒ g'(0) = 1/3
Given: h(g(g(x))) = x
⇒ h(g(g(f(x)))) = f(x)
⇒ h(g(x)) = f(x)
⇒ h(g(f(x))) = f(f(x))
⇒ h(x) = f(f(x))
At x = 0, h(0) = f(f(0)) ⇒ h(0) = f(2) ⇒ h(0) = 16
Now, h(x) = f(f(x))
Differentiating both sides,
h'(x) = f'(f(x))f'(x)
At x = 1, h'(1) = f'(f(1))f'(1) ⇒ h'(1) = f'(6) ⋅ 6 ⇒ h'(1) = 3 ⋅ 37 ⋅ 6 = 666
Now, h(g(x)) = f(x) ⇒ h(g(3)) = f(3) ⇒ h(g(3)) = 38