The function f(x) = frac{e^{2x} - 1}{e^{2x} + | vedclass

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

Question

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

Vedclass · Accepted Answer

Increasing

Vedclass · Answer

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

The function $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$ is a ........ function.

Similar Questions

Prove that the logarithmic function $f(x) = \log x$ is strictly increasing on $(0, \infty).$

In which interval is the function $f(x) = \frac{x}{\ln x}$ increasing?

For what values of $x$ is the function $f(x) = x^3 - 27x + 5$ strictly increasing?

The number of real roots of the equation $x^5+3x^3+4x+30=0$ is

Which one is the correct statement about the function $f(x) = \sin 2x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module