What type of function is f(x) = frac{e^{2x} - | vedclass

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

Question

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

Vedclass · Accepted Answer

Increasing

Vedclass · Answer

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

What type of function is $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$?

Similar Questions

The function $f$ defined by $f(x) = (x + 2)e^{-x}$ is

If $f(x) = x^3 + bx^2 + cx + d$ and $0 < b^2 < c$,then in $(-\infty, \infty)$:

Find the intervals in which the function $f$ given by $f(x)=2 x^{2}-3 x$ is
$(a)$ increasing
$(b)$ decreasing

Let $f(x) = \sin^4 x + \cos^4 x$. Then $f$ is an increasing function in the interval:

Prove that $y = \frac{4 \sin \theta}{2 + \cos \theta} - \theta$ is an increasing function of $\theta$ in the interval $\left[0, \frac{\pi}{2}\right].$

Vedclass Test Series

Exam Paper Generator

Online Exam Module