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Question Number 158455 by mnjuly1970 last updated on 04/Nov/21

Answered by MJS_new last updated on 05/Nov/21

$$\sqrt{x\lfloor x\rfloor +x^2\lfloor x-1\rfloor }=\sqrt{x((x+1)\lfloor x\rfloor -x)}$$$$g\left(x\right)=x((x+1)\lfloor x\rfloor -x)=0\iff x=0\vee (x+1)\lfloor x\rfloor =x$$$$\Rightarrow x=0\vee x=-\frac{1}{2}$$$$\mathrm{but}(\underset{x\to 1^{-}}{\lim }g\left(x\right)=-1)\ne (\underset{x\to 1^{+}}{\lim }g\left(x\right)=1)$$$$\Rightarrow f\left(x\right)\mathrm{defined}\mathrm{for}-\frac{1}{2}⩽x⩽0\vee x⩾1$$

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