Set, Relation and FunctionHard
Question
Let f: (-1, 1) → R be a differentiable function with f(0) = - 1 and f′(0) = 1. Let g(x) = [f(2f(x) + 2)]2. Then g′(0) =
Options
A.-4
B.0
C.-2
D.4
Solution
g′(x) = 2(f(2f(x) + 2)) 
= 2f(2f(x) + 2) f′(2f(x) + 2) . (2f′(x))
⇒ g′(0) = 2f(2f(0) + 2) . f′(2f(0) + 2) . 2(f′(0) = 4f(0) f′(0)
= 4(-1) (1) = - 4
= 2f(2f(x) + 2) f′(2f(x) + 2) . (2f′(x))⇒ g′(0) = 2f(2f(0) + 2) . f′(2f(0) + 2) . 2(f′(0) = 4f(0) f′(0)
= 4(-1) (1) = - 4
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