Given the function f(x) = left( frac{e^{2x} - | vedclass

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

Question

(A) Given $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.To check if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotien

Vedclass · Accepted Answer

Increasing

Vedclass · Answer

(A) Given $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$.To check if the function is increasing or decreasing,we find its derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.Let $u = e^{2x} - 1$ and $v = e^{2x} + 1$. Then $u' = 2e^{2x}$ and $v' = 2e^{2x}$.$f'(x) = \frac{(2e^{2x})(e^{2x} + 1) - (e^{2x} - 1)(2e^{2x})}{(e^{2x} + 1)^2}$$f'(x) = \frac{2e^{4x} + 2e^{2x} - (2e^{4x} - 2e^{2x})}{(e^{2x} + 1)^2}$$f'(x) = \frac{4e^{2x}}{(e^{2x} + 1)^2}$.Since $e^{2x} > 0$ for all $x \in \mathbb{R}$,$f'(x) > 0$ for all $x \in \mathbb{R}$.Therefore,$f(x)$ is an increasing function.

Given the function $f(x) = \left( \frac{e^{2x} - 1}{e^{2x} + 1} \right)$,the function is:

Similar Questions

What is the nature of the function $f(x) = \sin x$ at $x = 2\pi / 3$?

Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is $(i)$ increasing $(ii)$ decreasing.

The function $f(x) = \frac{\ln(\pi+x)}{\ln(e+x)}$ is

If $f(x) = \sin x - \cos x - ax + b$ is a decreasing function for all $x \in R$,then:

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module