Question Number 115882 by Khalmohmmad last updated on 29/Sep/20
Answered by PRITHWISH SEN 2 last updated on 29/Sep/20
![sgn(x+1)=1 when x>−1 sgn(x+1)=0 when x+1=0 sgn(x+1)=−1 when x<−1 ∵ denominator ≠0 ∴ sgn(x+1)≠ 1 ∴ domain x∈ (−∞,−1]](https://www.tinkutara.com/question/Q115938.png)
$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{1}\:\:\mathrm{when}\:\mathrm{x}>−\mathrm{1} \\ $$$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{0}\:\:\mathrm{when}\:\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)=−\mathrm{1}\:\mathrm{when}\:\mathrm{x}<−\mathrm{1} \\ $$$$\because\:\mathrm{denominator}\:\neq\mathrm{0}\:\therefore\:\mathrm{sgn}\left(\mathrm{x}+\mathrm{1}\right)\neq\:\mathrm{1} \\ $$$$\therefore\:\mathrm{domain}\:\mathrm{x}\in\:\left(−\infty,−\mathrm{1}\right] \\ $$