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

Find the derivative of the following function: $5 \sec x + 4 \cos x$.

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to

$\frac{d}{d x}\left[\left(x^{\frac{5}{2}}-x^{\frac{3}{2}}+1\right)\left(x^2-3 x+5\right)\right]=$

If $f(x) = 4x^3 + 3x^2 + 3x + 4$,$x \neq 0$,then $\frac{d}{dx}\left(x^3 \cdot f\left(\frac{1}{x}\right)\right) =$ . . . . . .

Differentiate the function with respect to $x$: $\cos (\sin x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module