Let f : R rightarrow R be defined as f(x) = x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = x^{3} + x - 5$.First,we find the derivative $f^{\prime}(x) = 3x^{2} + 1$.Since $f^{\prime}(x) > 0$ for all $x \in R$,the function $f

Vedclass · Accepted Answer

$\frac{1}{49}$

Vedclass · Answer

(A) Given $f(x) = x^{3} + x - 5$.First,we find the derivative $f^{\prime}(x) = 3x^{2} + 1$.Since $f^{\prime}(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing and therefore invertible.Given $f(g(x)) = x$,by the chain rule,we have $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$,which implies $g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.To find $g^{\prime}(63)$,we need to find $x$ such that $f(x) = 63$.$x^{3} + x - 5 = 63 \Rightarrow x^{3} + x - 68 = 0$.By inspection,for $x = 4$,$4^{3} + 4 - 5 = 64 + 4 - 5 = 63$. Thus,$g(63) = 4$.Now,$g^{\prime}(63) = \frac{1}{f^{\prime}(g(63))} = \frac{1}{f^{\prime}(4)}$.Since $f^{\prime}(x) = 3x^{2} + 1$,we have $f^{\prime}(4) = 3(4)^{2} + 1 = 3(16) + 1 = 48 + 1 = 49$.Therefore,$g^{\prime}(63) = \frac{1}{49}$.

Let $f : R \rightarrow R$ 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 $g^{\prime}(63)$ is equal to

Similar Questions

$\frac{d}{dx} [3 \sin(60^{\circ} - x^{\circ}) - 4 \cos^3(30^{\circ} + x^{\circ})] = \rule{1cm}{0.15mm}$

Find the derivative of the function: $4 \sqrt{x} - 2$. (Assume $a, b, c, d, p, q, r, s$ are fixed non-zero constants and $m, n$ are integers.)

If $y = \tan^{-1}(\frac{1}{x^2+x+1}) + \tan^{-1}(\frac{1}{x^2+3x+3}) + \tan^{-1}(\frac{1}{x^2+5x+7}) + \dots$ upto $n$ terms, then $y'(0) = $

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

If $y = \sec(\tan^{-1}x)$,then $\frac{dy}{dx}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module