Question Number 41765 by kunal1234523 last updated on 12/Aug/18
$${find}\:{range}\:{of}\:{the}\:{function}\:{f}\: \\ $$$${defined}\:{by}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$
Answered by ajfour last updated on 12/Aug/18
$${R}−\left[\mathrm{0},\mathrm{1}\right) \\ $$
Answered by alex041103 last updated on 12/Aug/18
$${lets}\:{define}\:{the}\:{inverse}\:{function}\:{of}\:{f}\left({x}\right) \\ $$$${as}\:{F}\left({f}\right).\:{Then}\:{the}\:{domain}\:{of}\:{F}\:{is}\:{the} \\ $$$${range}\:{of}\:{f}. \\ $$$${so}: \\ $$$${f}=\frac{\mathrm{1}}{\mathrm{1}−{F}^{\:\mathrm{2}} } \\ $$$$\Rightarrow{F}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{{f}}\:\Rightarrow\:{F}\left({f}\right)=\pm\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{f}}}. \\ $$$${The}\:{domain}\:{of}\:{F}\:{is}\:{defined}\:{by} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{f}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\geqslant\frac{\mathrm{1}}{{f}} \\ $$$${for}\:{f}<\mathrm{0}:\:\mathrm{1}>\frac{\mathrm{1}}{{f}}\:{as}\:\mathrm{1}>\mathrm{0}. \\ $$$${for}\:{f}>\mathrm{0}:{f}\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{range}\:{is}\:{f}\left({x}\right)\in\left(−\infty,\mathrm{0}\right)\cap\left[\mathrm{1},\infty\right) \\ $$
Commented by kunal1234523 last updated on 13/Aug/18
$${sir}\:{is}\:{it}\:\left(−\infty,\mathrm{0}\right)\cup\left[\mathrm{1},\infty\right) \\ $$
Commented by kunal1234523 last updated on 13/Aug/18
$$\left(−\infty,\mathrm{0}\right)\cap\left[\mathrm{1},\infty\right)\:=\:\phi\:,\:{i}\:{think} \\ $$
Commented by alex041103 last updated on 13/Aug/18
$${Oh}...{Yes}... \\ $$