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Question Number 114878 by bobhans last updated on 21/Sep/20

G i v e n f (x) = \underset{0}{\int^{x}} \frac{d t}{\sqrt{1 + t^{3}}} a n d g (x) b e t h e

i n v e r s e f u n c t i o n o f f (x), t h e n g^{″} (x) = λ g^{2} (x) .

t h e n t h e v a l u e o f λ =

Answered by Olaf last updated on 21/Sep/20

g o f (x) = x

f^{'} (x) \times g^{'} o f (x) = 1

g^{'} o f (x) = \frac{1}{f^{'} (x)} = \sqrt{1 + x^{3}}

f^{'} (x) \times g^{″} o f (x) = \frac{3 x^{2}}{2 \sqrt{1 + x^{3}}}

g^{″} o f (x) = \frac{1}{f^{'} (x)} [\frac{3 x^{2}}{2 \sqrt{1 + x^{3}}}]

g^{″} o f (x) = \sqrt{1 + x^{3}} [\frac{3 x^{2}}{2 \sqrt{1 + x^{3}}}]

g^{″} o f (x) = \frac{3}{2} x^{2}

g^{″} o f (x) = \frac{3}{2} {[g o f (x)]}^{2}

Let y = f (x)

g^{″} (y) = \frac{3}{2} g^{2} (y)

g^{″} (y) = λ g^{2} (y) with λ = \frac{3}{2}

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

f^{^{'}} (x) = \frac{1}{\sqrt{1 + x^{3}}} now, g^{^{'}} (x) = \frac{1}{f^{^{'}} {g (x)}}

g^{'} (x) = \sqrt{1 + {g (x)}^{3}}

g^{^{″}} (x) = \frac{1}{2 \sqrt{1 + {g (x)}^{3}}} . {3 g}^{2} (x) . g^{^{'}} (x)

λ g^{2} (x) = \frac{1}{2 g^{'} (x)} . 3 g^{2} (x) . g^{^{'}} (x)

λ = \frac{3}{2}