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

Given f(x) = ∫_0 ^x  (dt/( (√(1+t^3 )))) and g(x) be the  inverse function of f(x), then g ′′(x)=λg^2 (x).  then the value of λ =

Answered by Olaf last updated on 21/Sep/20

gof(x) = x  f′(x)×g′of(x) = 1  g′of(x) = (1/(f′(x))) = (√(1+x^3 ))  f′(x)×g′′of(x) = ((3x^2 )/(2(√(1+x^3 ))))  g′′of(x) = (1/(f′(x)))[((3x^2 )/(2(√(1+x^3 ))))]  g′′of(x) = (√(1+x^3 ))[((3x^2 )/(2(√(1+x^3 ))))]  g′′of(x) = (3/2)x^2   g′′of(x) = (3/2)[gof(x)]^2   Let y = f(x)  g′′(y) = (3/2)g^2 (y)  g′′(y) = λg^2 (y) with λ = (3/2)

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

f^′ (x)= (1/( (√(1+x^3 ))))     now, g^′ (x)= (1/(f^′ {g(x)}))  g′(x)= (√(1+{g(x)}^3 ))  g^(′′) (x)= (1/(2(√(1+{g(x)}^3 )))) . 3g^2 (x).g^′ (x)    λg^2 (x) = (1/(2g′(x))).3g^2 (x).g^′ (x)    λ= (3/2)