Let-f-be-a-real-valued-function-defined-on-the-interval-1-1-such-that-e-x-f-x-2-0-x-t-4-1-dt-x-1-1-and-let-g-be-the-inverse-function-of-f-Find-the-value-of-g-2-

Question Number 115260 by bobhans last updated on 24/Sep/20

L e t f b e a r e a l v a l u e d f u n c t i o n d e f i n e d

o n t h e i n t e r v a l (- 1, 1) s u c h t h a t

e^{- x} . f (x) = 2 + \underset{0}{\int^{x}} \sqrt{t^{4} + 1} d t \forall x \in (- 1, 1)

a n d l e t g b e t h e i n v e r s e f u n c t i o n o f f

. F i n d t h e v a l u e o f g^{'} (2) .

Commented by PRITHWISH SEN 2 last updated on 24/Sep/20

Answered by john santu last updated on 24/Sep/20

D i f f e r e n t i a t i n g g i v e n e q u a t i o n w e

g e t e^{- x} . f^{'} (x) - e^{- x} . f (x) = \sqrt{1 + x^{4}}

s i n c e (g \circ f) (x) = x a s g i s i n v e r s e o f f

\Rightarrow g (f (x)) = x \Rightarrow f^{'} (x) . g^{'} (f (x)) = 1

\Rightarrow g^{'} (f (0)) = \frac{1}{f^{'} (0)} \Rightarrow g^{'} (2) = \frac{1}{f^{'} (0)}

(h e r e f (0) = 2 o b t a i n e d f r o m

g i v e n e q u a t i o n .)

p u t x = 0 w e g e t f^{'} (0) = 3 .

∴ g^{'} (2) = \frac{1}{3}

Answered by mindispower last updated on 25/Sep/20

g^{'} (f (x)) = \frac{1}{f^{'} (x)}, f^{'} (x) = e^{x} (2 + \int_{0}^{x} \sqrt{1 + t^{4}}) + \sqrt{x^{4} + 1} e^{x})

g^{'} (f (0)) = \frac{1}{f^{'} (0)} = g^{'} (2) = \frac{1}{3}