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f-1-x-1-x-x-2-then-what-is-the-domain-and-range-of-f-1-x-




Question Number 103457 by bobhans last updated on 15/Jul/20
f(((1−x)/(1+x))) = x+2 then what is the domain  and range of f^(−1) (x) ?
$${f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\:=\:{x}+\mathrm{2}\:{then}\:{what}\:{is}\:{the}\:{domain} \\ $$$${and}\:{range}\:{of}\:{f}^{−\mathrm{1}} \left({x}\right)\:? \\ $$
Answered by Worm_Tail last updated on 15/Jul/20
   f(((1−x)/(1+x)))=x+2     (((1−x)/(1+x)))=f^(−1) (x+2)_(x=x−2)       ((3−x)/(−1+x))=f^(−1) (x)  Domain=R−{1}  3−x=−f^(−1) (x)+xf^(−1) (x)  x=((3+f^(−1) (x))/(1+f^(−1) (x)))     Range=R−{−1}
$$\:\:\:{f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}+\mathrm{2} \\ $$$$\:\:\:\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={f}^{−\mathrm{1}} \left({x}+\mathrm{2}\right)_{{x}={x}−\mathrm{2}} \\ $$$$\:\:\:\:\frac{\mathrm{3}−{x}}{−\mathrm{1}+{x}}={f}^{−\mathrm{1}} \left({x}\right)\:\:{Domain}={R}−\left\{\mathrm{1}\right\} \\ $$$$\mathrm{3}−{x}=−{f}^{−\mathrm{1}} \left({x}\right)+{xf}^{−\mathrm{1}} \left({x}\right) \\ $$$${x}=\frac{\mathrm{3}+{f}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}+{f}^{−\mathrm{1}} \left({x}\right)}\:\:\:\:\:{Range}={R}−\left\{−\mathrm{1}\right\} \\ $$
Answered by bemath last updated on 15/Jul/20
f^(−1) (x+2) = ((1−x)/(1+x))  f^(−1) (x) = ((1−(x−2))/(1+x−2)) = ((3−x)/(x−1))=((−x+3)/(x−1))  domain f^(−1) (x) ⇒ x∈R∧x≠1  for find Range f^(−1) (x)  ⇔x = ((f^(−1) (x)+3)/(f^(−1) (x)+1)) ; range  :y∈R∧y≠−1
$${f}^{−\mathrm{1}} \left({x}+\mathrm{2}\right)\:=\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}−\left({x}−\mathrm{2}\right)}{\mathrm{1}+{x}−\mathrm{2}}\:=\:\frac{\mathrm{3}−{x}}{{x}−\mathrm{1}}=\frac{−{x}+\mathrm{3}}{{x}−\mathrm{1}} \\ $$$${domain}\:{f}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow\:{x}\in\mathbb{R}\wedge{x}\neq\mathrm{1} \\ $$$${for}\:{find}\:{Range}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\Leftrightarrow{x}\:=\:\frac{{f}^{−\mathrm{1}} \left({x}\right)+\mathrm{3}}{{f}^{−\mathrm{1}} \left({x}\right)+\mathrm{1}}\:;\:{range} \\ $$$$:{y}\in\mathbb{R}\wedge{y}\neq−\mathrm{1} \\ $$

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