Question

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

• A. increasing
• B. Decreasing
• C. Neither increasing for decreasing
• D. Even function

Answer 1

The correct option is A increasing

Given f(x) = $\frac{e^{2x}-1}{e^{2x}+1}$

$f\left(x\right)=\frac{(e^{2x}+1)\frac{d}{dx}(e^{2x}-1)}{(e^{2x}+1{)}^2}$ $-(e^{2x}-1)\frac{d}{dx}(e^{2x}+1)$

$=\frac{\left(e^{2x+1}\right)\left(2e^{2x}\right)-(e^{2x}-1)\left(2e^{2x}\right)}{(e^{2x}+1{)}^2}$

$=\frac{4e^{2x}}{(e^{2x}+1{)}^2}>0$

$\therefore$ so the given function is increasing.