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Question Number 34736 by math khazana by abdo last updated on 10/May/18

$$letf\left(x\right)=-2x+\sqrt{x-3}$$$$1)find{f}^{-1}\left(x\right)$$$$2)calculate{\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right)and{\left(f^{-1}\right)}^{,}\left(2\right)$$$$3)letg\left(x\right)=x^2-2x+3$$$$calculatefog\left(x\right)andgive{D}_{fog}$$$$4)find{\left(fog\right)}^{-1}\left(x\right)$$$$5)calculate{\left({\left(fog\right)}^{-1}\right)}^{{}^{\prime }}\left(x\right).$$

Commented by prof Abdo imad last updated on 13/May/18

$$D_f=[-3,+\mathrm{\infty }[letputy=f\left(x\right)\iff x=f^{-1}\left(y\right)$$$$y=f\left(x\right)\iff y=-2x+\sqrt{x-3}\iff {(y+2x)}^2=x-3$$$$\Rightarrow y^2+4yx+4x^2-x+3=0\Rightarrow$$$$4x^2+(4y-1)x+y^2+3=0$$$$\mathrm{\Delta }={(4y-1)}^2-16(y^2+3)$$$$=16y^2-8y+1-16y^2-48=-8y-47$$$$=-8(y+\frac{47}{8})\mathrm{\Delta }mustbe⩾0\Rightarrow y⩽-\frac{47}{8}$$$$andy+2x⩾0x_1=\frac{1-4y+\sqrt{-8y-47}}{8}$$$$x_2=\frac{1-4y-\sqrt{-8y-47}}{8}$$$$y+2x_1=y+\frac{1-4y+\sqrt{-8y-47}}{4}$$$$=\frac{1+\sqrt{-8y-47}}{4}⩾0\Rightarrow f^{-1}\left(x\right)=\frac{1-4x+\sqrt{-8x-47}}{8}$$$$with{Df}^{-1}=]-\mathrm{\infty },-\frac{47}{8}]$$$$2){\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right)=-\frac{1}{2}+\frac{1}{8}\frac{-8}{2\sqrt{-8x-47}}$$$$=-\frac{1}{2}-\frac{1}{\sqrt{-8x-47}}but2\notin ]+\mathrm{\infty },-\frac{47}{8}[so$$$${\left(f^{-1}\right)}^{{}^{\prime }}\left(2\right)dontexist$$$$3)wehavef\left(x\right)=-2x+\sqrt{x-3}and$$$$g\left(x\right)=x^2-2x+3\Rightarrow$$$$fog\left(x\right)=-2g\left(x\right)+\sqrt{g\left(x\right)-3}$$$$=-2x^2+4x-6+\sqrt{x^2-2x+3-3}$$$$=-2x^2+4x-6+\sqrt{x^2-2x}$$$$D_{fog}=]-\mathrm{\infty },0]\cup [2,+\mathrm{\infty }[$$

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