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Question Number 159472 by mnjuly1970 last updated on 17/Nov/21

Answered by MJS_new last updated on 18/Nov/21

$$1-x^2⩾0\Rightarrow -1⩽x⩽1$$$$x+\lfloor x\rfloor ⩾0\Rightarrow x⩾0$$$$\sqrt{1-x^2}-\sqrt{x+\lfloor x\rfloor }⩾0$$$$1-x^2=x+\lfloor x\rfloor$$$$\Rightarrow x=-\frac{1+\sqrt{17}}{2}\vee x=\frac{-1+\sqrt{5}}{2}$$$$\Rightarrow$$$$f\left(x\right)\mathrm{defined}\mathrm{for}x=-\frac{1+\sqrt{17}}{2}\vee 0⩽x⩽\frac{-1+\sqrt{5}}{2}$$$$[{}_{\sqrt{1-x^2}-\sqrt{x+\lfloor x\rfloor }=0\Rightarrow f\left(x\right)=0}^{\mathrm{at}x=-\frac{1+\sqrt{17}}{2}\mathrm{both}\sqrt{1-x^2}\mathrm{and}\sqrt{x+\lfloor x\rfloor }\notin \mathbb{R}\mathrm{but}}$$$$\Rightarrow 0⩽f\left(x\right)⩽1$$

Commented by mnjuly1970 last updated on 19/Nov/21

$$merceymasterthankyousomuch$$

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