If the function f(x) = x^5 + e^{x/5} and g(x) | vedclass

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

Question

(A) Given $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$.We need to find the value of $\frac{1}{g'(1 + e^{1/5})}$.By the property of inverse functions,

Vedclass · Accepted Answer

$5 + \frac{e^{1/5}}{5}$

Vedclass · Answer

(A) Given $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$.We need to find the value of $\frac{1}{g'(1 + e^{1/5})}$.By the property of inverse functions,$g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$.Here,$y = 1 + e^{1/5}$.Setting $f(x) = x^5 + e^{x/5} = 1 + e^{1/5}$,we observe that $x = 1$ satisfies the equation because $1^5 + e^{1/5} = 1 + e^{1/5}$.Now,find the derivative $f'(x) = \frac{d}{dx}(x^5 + e^{x/5}) = 5x^4 + \frac{1}{5}e^{x/5}$.At $x = 1$,$f'(1) = 5(1)^4 + \frac{1}{5}e^{1/5} = 5 + \frac{e^{1/5}}{5}$.Since $g'(y) = \frac{1}{f'(x)}$,we have $g'(1 + e^{1/5}) = \frac{1}{f'(1)} = \frac{1}{5 + \frac{e^{1/5}}{5}}$.Therefore,$\frac{1}{g'(1 + e^{1/5})} = f'(1) = 5 + \frac{e^{1/5}}{5}$.

If the function $f(x) = x^5 + e^{x/5}$ and $g(x) = f^{-1}(x)$,then the value of $\frac{1}{g'(1 + e^{1/5})}$ is

Similar Questions

If $f:[1, \infty) \rightarrow [1, \infty)$ is defined by $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$, then $f^{-1}(3) =$

Let $f(x) = \int\limits_2^x \frac{dt}{\sqrt{1 + t^4}}$ and $g$ be the inverse of $f$. Then the value of $g'(0)$ is

Which of the following functions cannot have their inverse defined? (where $[.] \to$ greatest integer function)

Let $f(x) = \frac{x^2 - x}{x^2 + 2x}$,$x \ne 0, -2$. Then $\frac{d}{dx}[f^{-1}(x)]$ (wherever it is defined) is equal to

Let the function $f$ be defined by $f(x) = \frac{2x + 1}{1 - 3x}$,then $f^{-1}(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module