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Question Number 205680 by mnjuly1970 last updated on 27/Mar/24

$$$$$$solve$$$$$$$$\lfloor x\rfloor +\lfloor x^2\rfloor =\lfloor x^3\rfloor$$$$$$

Answered by Frix last updated on 27/Mar/24

$$\mathrm{Obviously}x=0$$$$-1<x<0:\lfloor x^3\rfloor =\lfloor x\rfloor =-1\wedge \lfloor x^2\rfloor =0$$$$0<x<1:\lfloor x^3\rfloor =\lfloor x^2\rfloor =\lfloor x\rfloor =0$$$$\Rightarrow$$$$-1<x<1★$$$$$$$$x=2^{\frac{1}{3}}:\lfloor x^3\rfloor =2\wedge \lfloor x^2\rfloor =\lfloor x\rfloor =1$$$$\mathrm{The}\mathrm{change}\mathrm{is}\mathrm{at}x=2^{\frac{1}{2}}:\lfloor x^2\rfloor =2$$$$\Rightarrow$$$$2^{\frac{1}{3}}⩽x<2^{\frac{1}{2}}★$$$$$$$$x=3^{\frac{1}{3}}:\lfloor x^3\rfloor =3\wedge \lfloor x^2\rfloor =2\wedge \lfloor x\rfloor =1$$$$\mathrm{The}\mathrm{change}\mathrm{is}\mathrm{at}x=4^{\frac{1}{3}}:\lfloor x^3\rfloor =4$$$$\Rightarrow$$$$3^{\frac{1}{3}}⩽x<4^{\frac{1}{3}}★$$

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