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

      f(x )= x + ⌊  x + (( ⌊  x_ ^   ⌋)/(⌊   (x^( 2) /(1 +x^( 2) ))   ⌋+1)) ⌋        ⇒  f^( −1)  (x )=?

$$ \\ $$$$\:\:\:\:{f}\left({x}\:\right)=\:{x}\:+\:\lfloor\:\:{x}\:+\:\frac{\:\lfloor\:\:\underset{} {\overset{} {{x}}}\:\:\rfloor}{\lfloor\:\:\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{1}\:+{x}^{\:\mathrm{2}} }\:\:\:\rfloor+\mathrm{1}}\:\rfloor \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{f}^{\:−\mathrm{1}} \:\left({x}\:\right)=? \\ $$

Commented by Frix last updated on 14/Jan/23

f(x)=x+2⌊x⌋  −2≤x<−1: f(x)=x−4∧−6≤f(x)<−5  −1≤x<0: f(x)=x−2∧−3≤f(x)<−2  0≤x<1: f(x)=x∧0≤f(x)<1  1≤x<2: f(x)=x+2∧3≤f(x)<4  2≤x<3: f(x)=x+4∧6≤f(x)<7    n≤x<n+1: f(x)=x+2n∧3n≤f(x)<3n+1∀n∈Z  ⇒  f^(−1) (x) is not defined for 3n−2≤x<3n∀n∈Z  f^(−1) (x)=x−2n; 3n≤x<3n+1∀n∈Z

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

Commented by mnjuly1970 last updated on 14/Jan/23

mercey sir frix

$${mercey}\:{sir}\:{frix} \\ $$

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