The interval(s), where the function $f(x) =\left\{\begin{matrix}\frac{1-e^x}{e^{2x}-1}:&x≠0\\\frac{-1}{2}:&x=0\end{matrix}\right.$ is increasing, is/are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $(-∞, ∞)$
For $x\ne 0$: $f(x)=\frac{1-e^x}{e^{2x}-1} =\frac{-(e^x-1)}{(e^x-1)(e^x+1)}=-\frac{1}{e^x+1}$, and $f(0)=-\frac{1}{2}$ matches this; hence $f(x)=-\frac{1}{e^x+1}$ for all $x\in\mathbb{R}$.
$f'(x)=\frac{e^x}{(e^x+1)^2}>0\ \ \forall x\in\mathbb{R}$.
Increasing on: $(-\infty,\infty)$.