Let-f-x-be-differentiable-function-and-g-x-be-the-inverse-of-f-x-when-lim-x-2-f-x-x-3-x-2-4-1-2-then-dg-x-dx-x-8-

Question Number 123752 by liberty last updated on 27/Nov/20

L e t f (x) b e d i f f e r e n t i a b l e f u n c t i o n

a n d g (x) b e t h e i n v e r s e o f f (x) .

w h e n \lim_{x \to 2} (\frac{f (x) - x^{3}}{x^{2} - 4}) = \frac{1}{2} t h e n \frac{d g (x)}{d x} ∣_{x = 8} = ?

Commented by benjo_mathlover last updated on 28/Nov/20

Answered by ajfour last updated on 27/Nov/20

s a y f (x) - x^{3} = 2 (x - 2)

\Rightarrow f (x) = x^{3} + 2 x - 4

x = g^{3} + 2 g - 4

f o r x = 8 g^{3} + 2 g - 12 = 0

\Rightarrow g = 2

1 = 3 g^{2} (\frac{d g}{d X}) + 2 (\frac{d g}{d x})

\Rightarrow (\frac{d g}{d x}) ∣_{x = 8} = \frac{1}{3 g^{2} + 2} = \frac{1}{14} .

Answered by bobhans last updated on 27/Nov/20

g (x) = f^{- 1} (x) t h e n g^{'} (x) = {(f^{- 1})}^{'} (x)

f r o m t h e l i m i t w e g e t f (2) = 8

o r f^{- 1} (8) = 2

w e k n o w t h a t g^{'} (8) = \frac{1}{f^{'} (f^{- 1} (8))} \dots (∙)

c o n s i d e r \lim_{x \to 2} \frac{f (x) - x^{3}}{x^{2} - 4} = \frac{1}{2}

\lim_{x \to 2} \frac{f^{'} (x) - 3 x^{2}}{2 x} = \frac{1}{2}

\Leftrightarrow f^{'} (2) - 12 = 2; f^{'} (2) = 14

s u b s t i t u t e i n t o (∙) w e g e t

g^{'} (8) = \frac{1}{f^{'} (f^{- 1} (8))} = \frac{1}{f^{'} (2)} = \frac{1}{14} .

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