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Question Number 129622 by Ar Brandon last updated on 16/Jan/21

$$\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^2\mathrm{for}\mathrm{all}\mathrm{x}⩾-1.\mathrm{If}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}\mathrm{the}$$$$\mathrm{function}\mathrm{whose}\mathrm{graph}\mathrm{is}\mathrm{the}\mathrm{reflection}\mathrm{of}\mathrm{the}\mathrm{graph}$$$$\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{y}=\mathrm{x},\mathrm{then}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}$$$$\mathrm{equal}\mathrm{to}...$$

Commented by mr W last updated on 16/Jan/21

$$g\left(x\right)=f^{-1}\left(x\right)=\sqrt{x}-1$$

Commented by Ar Brandon last updated on 17/Jan/21

Thanks Sir

Answered by bemath last updated on 17/Jan/21

$$\left(\begin{array}{c}{\mathrm{x}}^{\prime }\\ {\mathrm{y}}^{\prime }\end{array}\right)=\left(\begin{array}{c}01\\ 10\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)\Rightarrow \left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)={\left(\begin{array}{c}01\\ 10\end{array}\right)}^{-1}\left(\begin{array}{c}{\mathrm{x}}^{\prime }\\ {\mathrm{y}}^{\prime }\end{array}\right)$$$$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\frac{1}{0-1}\left(\begin{array}{c}0-1\\ -10\end{array}\right)\left(\begin{array}{c}{\mathrm{x}}^{\prime }\\ {\mathrm{y}}^{\prime }\end{array}\right)$$$$\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}01\\ 10\end{array}\right)\left(\begin{array}{c}{\mathrm{x}}^{\prime }\\ {\mathrm{y}}^{\prime }\end{array}\right)=\left(\begin{array}{c}{\mathrm{y}}^{\prime }\\ {\mathrm{x}}^{\prime }\end{array}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^2\Rightarrow \mathrm{y}={(\mathrm{x}+1)}^2$$$${\mathrm{x}}^{\prime }={({\mathrm{y}}^{\prime }+1)}^2;\mathrm{y}+1=\sqrt{\mathrm{x}};\mathrm{y}=\sqrt{\mathrm{x}}-1$$$$\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}-1.$$

Commented by Ar Brandon last updated on 17/Jan/21

Thanks bro

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