find-range-of-the-function-f-defined-by-f-x-1-1-x-2-

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(x) as F(f). Then the domain of F is the range of f. so: f=(1/(1−F^( 2) )) ⇒F^2 =1−(1/f) ⇒ F(f)=±(√(1−(1/f))). The domain of F is defined by 1−(1/f)≥0 ⇒1≥(1/f) for f<0: 1>(1/f) as 1>0. for f>0:f≥1 ⇒ range is f(x)∈(−∞,0)∩[1,∞)

$${lets}\:{define}\:{the}\:{inverse}\:{function}\:{of}\:{f}\left({x}\right) \\ $$$${as}\:{F}\left({f}\right).\:{Then}\:{the}\:{domain}\:{of}\:{F}\:{is}\:{the} \\ $$$${range}\:{of}\:{f}. \\ $$$${so}: \\ $$$${f}=\frac{\mathrm{1}}{\mathrm{1}−{F}^{\:\mathrm{2}} } \\ $$$$\Rightarrow{F}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{{f}}\:\Rightarrow\:{F}\left({f}\right)=\pm\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{f}}}. \\ $$$${The}\:{domain}\:{of}\:{F}\:{is}\:{defined}\:{by} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{f}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}\geqslant\frac{\mathrm{1}}{{f}} \\ $$$${for}\:{f}<\mathrm{0}:\:\mathrm{1}>\frac{\mathrm{1}}{{f}}\:{as}\:\mathrm{1}>\mathrm{0}. \\ $$$${for}\:{f}>\mathrm{0}:{f}\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{range}\:{is}\:{f}\left({x}\right)\in\left(−\infty,\mathrm{0}\right)\cap\left[\mathrm{1},\infty\right) \\ $$

Commented by kunal1234523 last updated on 13/Aug/18