Let f: R arrow R be defined as f ( x )= x 3+ x -5. If g(x) is a function such that f(g(x))=x, ∀ x ∈ R, then g prime(63) is equal to .

Question

Let f: R arrow R be defined as f ( x )= x 3+ x -5. If g(x) is a function such that f(g(x))=x, ∀ x ∈ R, then g prime(63) is equal to . (A) (1/49) (B) (3/49) (C) (43/49) (D) (91/49)

Tardigrade Admin · Accepted Answer

f(x)=x3+x-5 ⇒ f prime(x)=3 x2+1 ⇒ increasing function ⇒ invertible ⇒ g(x) is inverse of f(x) ⇒ g(f(x))=x ⇒ g prime(f(x)) f prime(x)=1 ⇒ f(x)=63 ⇒ x3+x-5=63 ⇒ x=4 put x=4 g prime(f(4)) f prime(4)=1 g prime(63) × 49=1 f prime(4)=49 g prime(63)=(1/49)

Q. Let $f : R \to R$ be defined as $f (x) = x^{3} + x - 5$ . If $g (x)$ is a function such that $f (g (x)) = x$ , $\forall x \in R$ , then $g^{'} (63)$ is equal to ______.

$\frac{1}{49}$

$\frac{3}{49}$

$\frac{43}{49}$

$\frac{91}{49}$

Q. Let f:R→R be defined as f(x)=x3+x−5. If g(x) is a function such that f(g(x))=x, ∀x∈R, then g′(63) is equal to ______.

491​

493​

4943​

4991​

f(x)=x3+x−5 ⇒f′(x)=3x2+1⇒ increasing function ⇒ invertible ⇒g(x) is inverse of f(x) ⇒g(f(x))=x ⇒g′(f(x))f′(x)=1 ⇒f(x)=63 ⇒x3+x−5=63 ⇒x=4 put x=4 g′(f(4))f′(4)=1 g′(63)×49=1{f′(4)=49} g′(63)=491​

Q. Let $f : R \to R$ be defined as $f (x) = x^{3} + x - 5$ . If $g (x)$ is a function such that $f (g (x)) = x$ , $\forall x \in R$ , then $g^{'} (63)$ is equal to ______.

$\frac{1}{49}$

$\frac{3}{49}$

$\frac{43}{49}$

$\frac{91}{49}$