Let f(x) = x5 + 2x3 + 3x + 1, x R, and g(x) be a function such that g(f(x)) = x for all x R. Then g(7)g'(7) is equal to :
- 42
- 7
- 1
- 14
Answer
(4) 14
Solution
g(f(x)) = x ⇒ g is the inverse of f
f(x) = 7 x5 + 2x3 + 3x = 6 x = 1 (only root, ∵ f'(x) > 0)
g(7) = 1
Differentiating: g'(f(x)) f'(x) = 1, f'(x) = 5x4 + 6x2 + 3
g'(7) = 1f'(1) = 114
g(7)g'(7) = 14
Answer: (4)