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

Let f(x)=(x+1)^2 for all x≥−1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y=x, then g(x) is equal to...

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

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

g(x)=f^(−1) (x)=(√x)−1

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

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

Thanks Sir

Answered by bemath last updated on 17/Jan/21

(((x′)),((y′)) ) = (((0 1)),((1 0)) ) ((x),(y) ) ⇒ ((x),(y) ) = (((0 1)),((1 0)) )^(−1) (((x′)),((y′)) ) ((x),(y) ) = (1/(0−1)) ((( 0 −1)),((−1 0)) ) (((x′)),((y′)) ) ((x),(y) ) = (((0 1)),((1 0)) ) (((x′)),((y′)) ) = (((y′)),((x′)) ) f(x)=(x+1)^2 ⇒y=(x+1)^2 x′ = (y′+1)^2 ; y+1 = (√x) ; y = (√x) −1 g(x)=(√x) −1 .

$$\:\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{0}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:\Rightarrow\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{0}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{0}−\mathrm{1}}\:\begin{pmatrix}{\:\:\:\:\mathrm{0}\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{0}}\end{pmatrix}\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{y}'}\\{\mathrm{x}'}\end{pmatrix} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow\mathrm{y}=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{x}'\:=\:\left(\mathrm{y}'+\mathrm{1}\right)^{\mathrm{2}} \:;\:\mathrm{y}+\mathrm{1}\:=\:\sqrt{\mathrm{x}}\:;\:\mathrm{y}\:=\:\sqrt{\mathrm{x}}\:−\mathrm{1} \\ $$$$\:\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\:−\mathrm{1}\:. \\ $$

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

Thanks bro

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