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Question Number 148911 by mathdanisur last updated on 01/Aug/21

$$f:[-3,0]\to [7,22]$$$$f\left(x\right)=x^2-2x+7$$$$find{f}^{-1}\left(x\right)=?$$

Answered by EDWIN88 last updated on 01/Aug/21

$$fisone-onesincef\left(x_1\right)=f\left(x_2\right)$$$$\Rightarrow x_1^2-2x_1+7=x_2^2-2x_2+7$$$$\Rightarrow (x_1-x_2)(x_1+x_2)-2(x_1-x_2)=0$$$$\Rightarrow (x_1-x_2)(x_1+x_2-2)=0$$$$\Rightarrow x_1=x_2[\because x_1+x_2-2\ne 0]$$$$lety\in Ythenfbeingontothere$$$$existxsuchthaty=f\left(x\right)$$$$Nowy=f\left(x\right)=x^2-2x+7$$$$y={(x-1)}^2+6$$$$\Rightarrow x=1+\sqrt{y-6}$$$$\Rightarrow f^{-1}\left(y\right)=1+\sqrt{y-6}$$$$thuswedefine{f}^{-1}\left(x\right)=1+\sqrt{x-6}$$

Commented by mathdanisur last updated on 01/Aug/21

$$ThankyouSer$$

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