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f-3-0-7-22-f-x-x-2-2x-7-find-f-1-x-

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Question Number 148911 by mathdanisur last updated on 01/Aug/21
f:[−3, 0]→[7, 22] f(x) = x^2 - 2x + 7 find f^( −1) (x) = ?
f:[−3,0]→[7,22]f(x)=x2−2x+7findf−1(x)=?
Answered by EDWIN88 last updated on 01/Aug/21
f is one−one since f(x_1 )=f(x_2 ) ⇒x_1 ^2 −2x_1 +7=x_2 ^2 −2x_2 +7 ⇒(x_1 −x_2 )(x_1 +x_2 )−2(x_1 −x_2 )=0 ⇒(x_1 −x_2 )(x_1 +x_2 −2)=0 ⇒x_1 =x_2 [ ∵ x_1 +x_2 −2 ≠ 0 ] let y∈Y then f being onto there exist x such that y=f(x) Now y=f(x)=x^2 −2x+7 y=(x−1)^2 +6 ⇒x=1+(√(y−6)) ⇒f^(−1) (y)=1+(√(y−6)) thus we define f^(−1) (x)=1+(√(x−6))
fisone−onesincef(x1)=f(x2)⇒x12−2x1+7=x22−2x2+7⇒(x1−x2)(x1+x2)−2(x1−x2)=0⇒(x1−x2)(x1+x2−2)=0⇒x1=x2[∵x1+x2−2≠0]lety∈Ythenfbeingontothereexistxsuchthaty=f(x)Nowy=f(x)=x2−2x+7y=(x−1)2+6⇒x=1+y−6⇒f−1(y)=1+y−6thuswedefinef−1(x)=1+x−6
Commented by mathdanisur last updated on 01/Aug/21
Thank you Ser
ThankyouSer

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