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) = ?](https://www.tinkutara.com/question/Q148911.png)
$${f}:\left[−\mathrm{3},\:\mathrm{0}\right]\rightarrow\left[\mathrm{7},\:\mathrm{22}\right] \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:-\:\mathrm{2}{x}\:+\:\mathrm{7} \\ $$$${find}\:\:\:{f}^{\:−\mathrm{1}} \left({x}\right)\:=\:? \\ $$
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))](https://www.tinkutara.com/question/Q148934.png)
$${f}\:{is}\:{one}−{one}\:{since}\:{f}\left({x}_{\mathrm{1}} \right)={f}\left({x}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{x}_{\mathrm{1}} +\mathrm{7}={x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{x}_{\mathrm{2}} +\mathrm{7} \\ $$$$\Rightarrow\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)−\mathrm{2}\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\left[\:\because\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} −\mathrm{2}\:\neq\:\mathrm{0}\:\right] \\ $$$${let}\:{y}\in{Y}\:{then}\:{f}\:{being}\:{onto}\:{there} \\ $$$${exist}\:{x}\:{such}\:{that}\:{y}={f}\left({x}\right) \\ $$$${Now}\:{y}={f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7} \\ $$$${y}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6}\: \\ $$$$\Rightarrow{x}=\mathrm{1}+\sqrt{{y}−\mathrm{6}}\: \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({y}\right)=\mathrm{1}+\sqrt{{y}−\mathrm{6}} \\ $$$${thus}\:{we}\:{define}\:{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{1}+\sqrt{{x}−\mathrm{6}} \\ $$
Commented by mathdanisur last updated on 01/Aug/21
$${Thank}\:{you}\:{Ser} \\ $$