The interval(s), where the function \(f(x) = \begin{cases}\displaystyle\frac{1-e^x}{e^{2x}-1} & : x \ne 0\\ \displaystyle\frac{-1}{2} & : x = 0\end{cases}\) is increasing, is/are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-\infty, \infty)\)
For x ≠ 0, the denominator is a difference of squares, \(e^{2x} - 1 = (e^x - 1)(e^x + 1)\):
\( \displaystyle f(x) = \frac{-(e^x - 1)}{(e^x - 1)(e^x + 1)} \)
\( \displaystyle f(x) = -\frac{1}{e^x + 1} \)
At x = 0 this formula gives \(-\frac{1}{1 + 1} = -\frac{1}{2}\), which is exactly the given f(0). So \(f(x) = -\frac{1}{e^x + 1}\) for every real x, with no break at 0.
\( \displaystyle f'(x) = \frac{e^x}{(e^x + 1)^2} \gt 0 \text{ for all } x \)
So f is increasing on \((-\infty, \infty)\); the point 0 does not have to be left out, as in Option (4).