Let f: R arrow R be defined as f(x)=x3+3 x+2 and g(x) be a function such that g(x)=x f-1(x)-∫ limits0f-1(x) f(t) d t, then

Question

Let f: R arrow R be defined as f(x)=x3+3 x+2 and g(x) be a function such that g(x)=x f-1(x)-∫ limits0f-1(x) f(t) d t, then (A) g prime(6)=1 (B) g prime(6)=(1/6) (C) g prime prime(6)=1 (D) g prime prime(6)=(1/6)

Tardigrade Admin · Accepted Answer

Clearly, f ( x ) is bijective, ∴ invertible Θ g(x)=x f-1(x)-∫ limits0f-1(x) f(t) d t Differentiation both sides g prime( x )=1 ⋅ f -1( x )+ x ( f -1( x )) prime- f ( f -1( x )) ⋅( f -1( x )) prime ⇒ g prime( x )= f -1( x ) ⇒ g prime(6)= f -1(6)=1 text and g prime prime( x )=( f -1) prime( x ) ⇒ g prime prime(6)=( f -1) prime(6)=(1/ f prime(1)) ⇒ g prime prime(6)=(1/6)

Q. Let $f : R \to R$ be defined as $f (x) = x^{3} + 3 x + 2$ and $g (x)$ be a function such that $g (x) = x f^{- 1} (x) - 0 \int f^{- 1} (x) f (t) d t$ , then

$g^{'} (6) = 1$

$g^{'} (6) = \frac{1}{6}$

$g^{''} (6) = 1$

$g^{''} (6) = \frac{1}{6}$

Q. Let f:R→R be defined as f(x)=x3+3x+2 and g(x) be a function such that g(x)=xf−1(x)−0∫f−1(x)​f(t)dt, then

g′(6)=1

g′(6)=61​

g′′(6)=1

g′′(6)=61​

Q. Let $f : R \to R$ be defined as $f (x) = x^{3} + 3 x + 2$ and $g (x)$ be a function such that $g (x) = x f^{- 1} (x) - 0 \int f^{- 1} (x) f (t) d t$ , then

$g^{'} (6) = 1$

$g^{'} (6) = \frac{1}{6}$

$g^{''} (6) = 1$

$g^{''} (6) = \frac{1}{6}$