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

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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(1−x1+x)=x+2thenwhatisthedomainandrangeoff−1(x)?
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(1−x1+x)=x+2(1−x1+x)=f−1(x+2)x=x−23−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}
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−1(x+2)=1−x1+xf−1(x)=1−(x−2)1+x−2=3−xx−1=−x+3x−1domainf−1(x)⇒x∈R∧x≠1forfindRangef−1(x)⇔x=f−1(x)+3f−1(x)+1;range:y∈R∧y≠−1

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