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Question Number 184942 by mnjuly1970 last updated on 14/Jan/23

$$$$$$f\left(x\right)=x+\lfloor x+\frac{\lfloor \underset{}{\stackrel{}{x}}\rfloor }{\lfloor \frac{x^2}{1+x^2}\rfloor +1}\rfloor$$$$\Rightarrow f^{-1}\left(x\right)=?$$

Commented by Frix last updated on 14/Jan/23

$$f\left(x\right)=x+2\lfloor x\rfloor$$$$-2⩽x<-1:f\left(x\right)=x-4\wedge -6⩽f\left(x\right)<-5$$$$-1⩽x<0:f\left(x\right)=x-2\wedge -3⩽f\left(x\right)<-2$$$$0⩽x<1:f\left(x\right)=x\wedge 0⩽f\left(x\right)<1$$$$1⩽x<2:f\left(x\right)=x+2\wedge 3⩽f\left(x\right)<4$$$$2⩽x<3:f\left(x\right)=x+4\wedge 6⩽f\left(x\right)<7$$$$$$$$n⩽x<n+1:f\left(x\right)=x+2n\wedge 3n⩽f\left(x\right)<3n+1\mathrm{\forall }n\in \mathbb{Z}$$$$\Rightarrow$$$$f^{-1}\left(x\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}3n-2⩽x<3n\mathrm{\forall }n\in \mathbb{Z}$$$$f^{-1}\left(x\right)=x-2n;3n⩽x<3n+1\mathrm{\forall }n\in \mathbb{Z}$$

Commented by mnjuly1970 last updated on 14/Jan/23

$$merceysirfrix$$

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