Question-28858

Question Number 28858 by amit96 last updated on 31/Jan/18

Commented by abdo imad last updated on 31/Jan/18

w e h a v e f (x) = x^{3} + x a n d g (x) = x^{3} - x \Rightarrow

f^{^{'}} (x) = 3 x^{2} + 1 a n d g^{^{'}} (x) = 3 x^{2} - 1 a n d

{({g o f}^{- 1})}^{^{'}} (2) = g^{^{'}} (f^{- 1} (2)) . f^{- 1^{^{'}}} (2) b u t f^{- 1} (2) = t \Leftrightarrow f (t) = 2

\Leftrightarrow t^{3} + t - 2 = 0 \Leftrightarrow (t - 1) (t^{2} + t + 2) = 0 \Leftrightarrow t = 1 b e c a u s e

t h e p o l y n o m i a l t^{2} + t + 2 h a v e n t a n y r o o t s (Δ < 0) s o

f^{- 1} (2) = 1 a n d {(f^{- 1})}^{^{'}} (2) = \frac{1}{f^{^{'}} (f^{- 1} (2))} = \frac{1}{f^{^{'}} (1)} = \frac{1}{4}

g^{^{'}} (f^{- 1} (2)) = g^{^{'}} (1) = 2 f o r t h a t {({g o f}^{- 1})}^{^{'}} (2) = 2 \times \frac{1}{4} = \frac{1}{2} (B)