Question

Let f : (–1, 1) → R be a differentiable function with f(0) = –1 and f ′(0) = 1, g(x) = [f(2f(x) + 2)]². Then g′(0) =

(a) 4 

(b) – 4 

(c) 0 

(d) –2

Answer 1

(b) : g(x) = {f(2f (x) + 2}²
We have on differentiation with respect to x, g′(x) = 2f (2f(x) + 2) · f ′(2f(x) + 2) · 2f ′(x)
Let x = 0
g′(0) = 2f(2f(0) + 2) · f ′(2f(0) + 2) · 2f ′(0)
= 2f (0) · f ′(0) · 2f ′(0) = (–2)(1)(2) = – 4.