Question Number 192852 by York12 last updated on 29/May/23
$$ \\ $$$${find}\:{the}\:{domain}\:{of}\:{thefunction} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\left\{{x}\right\}^{\mathrm{2}} }}\:\:\:\:\:{where}\:\left\{.\right\}\:{is}\:{the}\:{fractional}\:{part}\:{function}. \\ $$
Commented by York12 last updated on 30/May/23
$${the}\:{answer}\:{is} \\ $$$$\left(−\infty−\mathrm{1}\right)\:\cup\:\left[−\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\cup\:\left[\mathrm{1}\:,\:\infty\:\right) \\ $$$$ \\ $$
Commented by MM42 last updated on 30/May/23
![(−∞,−1)∪[−1,−(1/2)]∪[1,∞)=(−∞,−(1/2)]∪⌈1,∞)](https://www.tinkutara.com/question/Q192873.png)
$$\left.\left(−\infty,−\mathrm{1}\right)\cup\left[−\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\right]\cup\left[\mathrm{1},\infty\right)=\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right]\cup\lceil\mathrm{1},\infty\right) \\ $$
Answered by MM42 last updated on 29/May/23
$${D}=\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left[\mathrm{1},\infty\right) \\ $$
Commented by York12 last updated on 30/May/23
$$ \\ $$$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{define}\:\left\{{x}\right\}\:\mathrm{for}\:{x}<\mathrm{0}? \\ $$$$\left\{−\mathrm{1}.\mathrm{3}\right\}=−.\mathrm{3}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{learned}\:\mathrm{but}\:\mathrm{I}'\mathrm{ve} \\ $$$$\mathrm{also}\:\mathrm{seen}\:\left\{−\mathrm{1}.\mathrm{3}\right\}=.\mathrm{7} \\ $$$$\left\{{x}\right\}={x}−\lfloor{x}\rfloor \\ $$$${for}\:{example}\:{x}=−\mathrm{3}.\mathrm{4} \\ $$$$\left\{−\mathrm{3}.\mathrm{5}\right\}=\:−\mathrm{3}.\mathrm{4}\:−\:\lfloor−\mathrm{3}.\mathrm{4}\rfloor\:=\:−\mathrm{3}.\mathrm{4}\:+\mathrm{4}\:=\:.\mathrm{6} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MM42 last updated on 30/May/23
![In authoritative books and mathematical references ,the following difination is use correct part. [x]=x−{x} ; 0≤{x}<1 therfore always 0≤ {x}<1](https://www.tinkutara.com/question/Q192872.png)
$$\mathcal{I}{n}\:{authoritative}\:{books}\:{and}\:{mathematical}\:{references}\:,{the}\:{following}\:{difination}\:{is}\:{use}\: \\ $$$${correct}\:{part}.\:\left[{x}\right]={x}−\left\{{x}\right\}\:\:\:;\:\:\:\:\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1} \\ $$$${therfore}\:\:{always}\:\:\:\mathrm{0}\leqslant\:\left\{{x}\right\}<\mathrm{1} \\ $$
Commented by York12 last updated on 30/May/23
$${yeah}\:{sir}\:{exactly}\: \\ $$
Commented by York12 last updated on 30/May/23
![(−∞,−1)∪[−1,−(1/2)]∪[1,∞)=(−∞,−(1/2)]∪⌈1,∞) thanks sir yeah you are right](https://www.tinkutara.com/question/Q192877.png)
$$ \\ $$$$\left.\left(−\infty,−\mathrm{1}\right)\cup\left[−\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\right]\cup\left[\mathrm{1},\infty\right)=\left(−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\right]\cup\lceil\mathrm{1},\infty\right) \\ $$$${thanks}\:{sir}\:{yeah}\:{you}\:{are}\:{right} \\ $$$$ \\ $$$$ \\ $$
Commented by MM42 last updated on 30/May/23
$${good}\:{luck} \\ $$
Answered by witcher3 last updated on 31/May/23
![x^2 >{x}^2 x=[x]+{x} if ∣x∣≥1⇒x^2 −1≥x^2 −{x}^2 >0 {x}∈[0,1[ if x∈]−1,1[ x∈]−1,0[ x=−1+{x}⇒ x^2 −{x}^2 =1−2{x}>0⇒{x}<(1/2) ⇒x<−(1/2)⇒x∈]−1,−(1/2)[ if 1>x≥0 ⇒x={x}⇒x−{x}=0 ⇒D_f =]−∞,−(1/2)[∪[1,∞[](https://www.tinkutara.com/question/Q192922.png)
$$\mathrm{x}^{\mathrm{2}} >\left\{\mathrm{x}\right\}^{\mathrm{2}} \\ $$$$\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\} \\ $$$$\mathrm{if}\:\mid\mathrm{x}\mid\geqslant\mathrm{1}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{x}^{\mathrm{2}} −\left\{\mathrm{x}\right\}^{\mathrm{2}} >\mathrm{0} \\ $$$$\left\{\mathrm{x}\right\}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$$\left.\mathrm{if}\:\mathrm{x}\in\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\left.\mathrm{x}\in\right]−\mathrm{1},\mathrm{0}\left[\right. \\ $$$$\mathrm{x}=−\mathrm{1}+\left\{\mathrm{x}\right\}\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} −\left\{\mathrm{x}\right\}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}\left\{\mathrm{x}\right\}>\mathrm{0}\Rightarrow\left\{\mathrm{x}\right\}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\Rightarrow\mathrm{x}<−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}\in\right]−\mathrm{1},−\frac{\mathrm{1}}{\mathrm{2}}\left[\right. \\ $$$$\mathrm{if}\:\mathrm{1}>\mathrm{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\left\{\mathrm{x}\right\}\Rightarrow\mathrm{x}−\left\{\mathrm{x}\right\}=\mathrm{0} \\ $$$$\left.\Rightarrow\mathrm{D}_{\mathrm{f}} =\right]−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\left[\cup\left[\mathrm{1},\infty\left[\right.\right.\right. \\ $$