Question Number 206328 by mnjuly1970 last updated on 12/Apr/24
$$ \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:\lfloor\:\frac{\:\mathrm{1}+\:{x}\:+{x}^{\mathrm{2}} }{{x}+\:{x}^{\:\mathrm{2}} }\:\rfloor\:{is}\:{given}. \\ $$$$\:\:\:\:\:\:\:\Rightarrow\begin{cases}{\:\:{D}_{{f}} \:=\:?\:\left({domain}\:\right)}\\{\:\:\:{R}_{\:{f}} \:=\:?\left(\:{range}\right)}\end{cases} \\ $$$$ \\ $$
Answered by TheHoneyCat last updated on 12/Apr/24
![D_(⌊•⌋) =R So we need ((1+x+x^2 )/(x+x^2 ))=(1/(x+x^2 ))+1∈R So we need (1/x)×(1/(1+x))∈R This works on all R\{0,−1} buy we could also add {+∞,−∞} and also {x∈C ∣ Arg(x)=−Arg(1+x)} This boils down to: ∣x∣e^(iθ) (1+∣x∣e^(iθ) )∈R ⇒e^(iθ) +∣x∣e^(i2θ) ∈R ⇒sinθ+∣x∣sin(2θ)=0 But assuming you just work with real numbers: D_f = R\{−1,0} = ]−∞,−1[∪]−1,0[∪]0,∞[ R_f =⌊R_((1/(x+x^2 ))+1) ⌋ =⌊R_(1/(x+x^2 )) +1⌋ =⌊(R_(x+x^2 ∣ x∈R\{−1,0}) )^(−1) ⌋+1 Now, x+x^2 is a polynomial of deg 2 it has its minimu reached at ((−1)/2) ie R_(x+x^2 ) =[((−1)/4),+∞[ so: R_(x+x^2 ∣ x∈R\{−1,0}) =[−1/4,0[∪R_+ ^∗ so: (R_(x+x^2 ∣ x∈R\{−1,0}) )^(−1) =]−∞,−4]∪R_+ ^∗ Thus: R_f =(⌊ ]−∞,−4] ⌋+1)∪( ⌊R_+ ^∗ ⌋+1) =(⌊R_− ⌋−4+1)∪( Z_+ +1) =∣] −∞, −3 ∣] ∪ [∣1,∞[∣ { ((D_f =R\{−1,0})),((R_f =Z\{−2,−1,0} _□ )) :}](https://www.tinkutara.com/question/Q206333.png)
$${D}_{\lfloor\bullet\rfloor} =\mathbb{R} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{need}\:\frac{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{{x}+{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}+{x}^{\mathrm{2}} }+\mathrm{1}\in\mathbb{R} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{need}\:\frac{\mathrm{1}}{{x}}×\frac{\mathrm{1}}{\mathrm{1}+{x}}\in\mathbb{R} \\ $$$$\mathrm{This}\:\mathrm{works}\:\mathrm{on}\:\mathrm{all}\:\mathbb{R}\backslash\left\{\mathrm{0},−\mathrm{1}\right\}\: \\ $$$$\mathrm{buy}\:\mathrm{we}\:\mathrm{could}\:\mathrm{also}\:\mathrm{add}\:\left\{+\infty,−\infty\right\} \\ $$$$\mathrm{and}\:\mathrm{also}\:\left\{{x}\in\mathbb{C}\:\mid\:\mathrm{Arg}\left({x}\right)=−\mathrm{Arg}\left(\mathrm{1}+{x}\right)\right\} \\ $$$$\mathrm{This}\:\mathrm{boils}\:\mathrm{down}\:\mathrm{to}:\:\mid{x}\mid{e}^{{i}\theta} \left(\mathrm{1}+\mid{x}\mid{e}^{{i}\theta} \right)\in\mathbb{R} \\ $$$$\Rightarrow{e}^{{i}\theta} +\mid{x}\mid{e}^{{i}\mathrm{2}\theta} \in\mathbb{R} \\ $$$$\Rightarrow\mathrm{sin}\theta+\mid{x}\mid\mathrm{sin}\left(\mathrm{2}\theta\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{assuming}\:\mathrm{you}\:\mathrm{just}\:\mathrm{work}\:\mathrm{with}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\left.{D}_{{f}} =\:\mathbb{R}\backslash\left\{−\mathrm{1},\mathrm{0}\right\}\:\:=\:\right]−\infty,−\mathrm{1}\left[\cup\right]−\mathrm{1},\mathrm{0}\left[\cup\right]\mathrm{0},\infty\left[\right. \\ $$$$ \\ $$$${R}_{{f}} =\lfloor{R}_{\frac{\mathrm{1}}{{x}+{x}^{\mathrm{2}} }+\mathrm{1}} \rfloor \\ $$$$\:\:\:\:\:\:=\lfloor{R}_{\frac{\mathrm{1}}{{x}+{x}^{\mathrm{2}} }} +\mathrm{1}\rfloor \\ $$$$\:\:\:\:\:\:=\lfloor\left({R}_{{x}+{x}^{\mathrm{2}} \mid\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1},\mathrm{0}\right\}} \right)^{−\mathrm{1}} \rfloor+\mathrm{1} \\ $$$$\mathrm{Now},\:{x}+{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\mathrm{deg}\:\mathrm{2} \\ $$$$\mathrm{it}\:\mathrm{has}\:\mathrm{its}\:\mathrm{minimu}\:\mathrm{reached}\:\mathrm{at}\:\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{ie}\:{R}_{{x}+{x}^{\mathrm{2}} } =\left[\frac{−\mathrm{1}}{\mathrm{4}},+\infty\left[\right.\right. \\ $$$$\mathrm{so}: \\ $$$${R}_{{x}+{x}^{\mathrm{2}} \mid\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1},\mathrm{0}\right\}} =\left[−\mathrm{1}/\mathrm{4},\mathrm{0}\left[\cup\mathbb{R}_{+} ^{\ast} \right.\right. \\ $$$$\mathrm{so}: \\ $$$$\left.\left(\left.{R}_{{x}+{x}^{\mathrm{2}} \mid\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1},\mathrm{0}\right\}} \right)^{−\mathrm{1}} =\right]−\infty,−\mathrm{4}\right]\cup\mathbb{R}_{+} ^{\ast} \\ $$$$\mathrm{Thus}: \\ $$$$\left.{R}_{{f}} \left.=\left(\lfloor\:\:\right]−\infty,−\mathrm{4}\right]\:\:\rfloor+\mathrm{1}\right)\cup\left(\:\lfloor\mathbb{R}_{+} ^{\ast} \rfloor+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:=\left(\lfloor\mathbb{R}_{−} \rfloor−\mathrm{4}+\mathrm{1}\right)\cup\left(\:\mathbb{Z}_{+} +\mathrm{1}\right) \\ $$$$\left.\:\left.\:\:\:\:\:=\mid\right]\:−\infty,\:−\mathrm{3}\:\mid\right]\:\:\cup\:\:\left[\mid\mathrm{1},\infty\left[\mid\right.\right. \\ $$$$\begin{cases}{{D}_{{f}} =\mathbb{R}\backslash\left\{−\mathrm{1},\mathrm{0}\right\}}\\{{R}_{{f}} =\mathbb{Z}\backslash\left\{−\mathrm{2},−\mathrm{1},\mathrm{0}\right\}\:\:\:\:_{\Box} }\end{cases} \\ $$
Commented by mr W last updated on 12/Apr/24
$${for}\:{which}\:{x}\:{such}\:{that}\:{f}\left({x}\right)=−\mathrm{3}? \\ $$
Commented by TheHoneyCat last updated on 12/Apr/24
$${f}\left(−\mathrm{1}/\mathrm{2}\right)=\lfloor\frac{\mathrm{1}}{\left(−\mathrm{1}/\mathrm{2}\right)+\left(−\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{1}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lfloor\frac{\mathrm{1}}{−\mathrm{1}/\mathrm{2}+\mathrm{1}/\mathrm{4}}+\mathrm{1}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lfloor\frac{\mathrm{1}}{−\mathrm{1}/\mathrm{4}}+\mathrm{1}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lfloor−\mathrm{4}+\mathrm{1}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lfloor−\mathrm{3}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{3} \\ $$$$\mathrm{But}\:\mathrm{since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{talking}\:\mathrm{about}\:\mathrm{it}, \\ $$$$\mathrm{notice}\:\mathrm{that}\:\mathrm{this}\:\mathrm{point}\:\mathrm{is}\:\mathrm{very}\:\mathrm{special}. \\ $$$$\mathrm{Most}\:\mathrm{points}\:{x}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{neigboorhood}. \\ $$$$\mathrm{meaning}\:{x}\pm\epsilon\:\mathrm{with}\:\mathrm{epsilon}\:\mathrm{tiny}\:\mathrm{enough}\: \\ $$$$\mathrm{will}\:\mathrm{give}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}. \\ $$$$\mathrm{some}\:\mathrm{points}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\mathrm{this}\:\mathrm{holds}\:\mathrm{only} \\ $$$$\mathrm{for}\:{x}+\epsilon\:\mathrm{or}\:{x}−\epsilon\:.\:\left(\mathrm{points}\:\mathrm{where}\:\mathrm{the}\:\lfloor\bullet\rfloor\:\mathrm{causes}\right. \\ $$$$\left.\mathrm{a}\:\mathrm{jump}\right) \\ $$$$\mathrm{But}\:{x}_{\mathrm{0}} =−\mathrm{1}/\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{point} \\ $$$$\mathrm{that}\:\mathrm{has}\:\mathrm{a}\:\mathrm{compleet}\:\mathrm{neigboorhood}\:\mathrm{that}\:\mathrm{is}\:\mathrm{worth} \\ $$$$−\mathrm{4}\:\mathrm{and}\:\mathrm{never}\:\mathrm{itself}. \\ $$$$ \\ $$$$\mathrm{this}\:\mathrm{means}\:\mathrm{lim}_{{x}\rightarrow{x}_{\mathrm{0}} } {f}\left({x}\right)=−\mathrm{4}\neq{f}\left({x}_{\mathrm{0}} \right) \\ $$$$ \\ $$$$\mathrm{In}\:\mathrm{more}\:\mathrm{fancy}\:\mathrm{words},\:{f}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{by}\:\mathrm{part} \\ $$$$\mathrm{everywhere}\:\mathrm{except}\:\mathrm{at}\:{x}_{\mathrm{0}} . \\ $$
Commented by mr W last updated on 12/Apr/24
$${great}! \\ $$
Answered by MM42 last updated on 12/Apr/24
![f(x)=[(1/(x^2 +x))]+1 if g(x)=(1/(x^2 +x))⇒f(x)=[g(x)]+1 D_f =D_g =R−{0,1} x→ { ((−1^− ⇒g→+∞ )),((−1^+ ⇒g→−∞)) :} & x→ { ((0^− ⇒g→−∞)),((0^+ ⇒g→+∞)) :} g′(x)=((−2x−1)/((x^2 +x)^2 )) ⇒M(−(1/2) , −4) ⇒R_g =(−∞,−4]∪(0,∞) R_f =Z−{−3,−2,−1}](https://www.tinkutara.com/question/Q206347.png)
$${f}\left({x}\right)=\left[\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}}\right]+\mathrm{1} \\ $$$${if}\:\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}}\Rightarrow{f}\left({x}\right)=\left[{g}\left({x}\right)\right]+\mathrm{1}\:\: \\ $$$${D}_{{f}} ={D}_{{g}} =\mathbb{R}−\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$${x}\rightarrow\begin{cases}{−\mathrm{1}^{−} \Rightarrow{g}\rightarrow+\infty\:\:\:}\\{−\mathrm{1}^{+} \Rightarrow{g}\rightarrow−\infty}\end{cases}\:\:\:\&\:\:\:\:{x}\rightarrow\begin{cases}{\mathrm{0}^{−} \Rightarrow{g}\rightarrow−\infty}\\{\mathrm{0}^{+} \Rightarrow{g}\rightarrow+\infty}\end{cases} \\ $$$${g}'\left({x}\right)=\frac{−\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}\right)^{\mathrm{2}} }\:\:\:\Rightarrow{M}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:,\:−\mathrm{4}\right) \\ $$$$\Rightarrow{R}_{{g}} =\left(−\infty,−\mathrm{4}\right]\cup\left(\mathrm{0},\infty\right) \\ $$$${R}_{{f}} =\mathbb{Z}−\left\{−\mathrm{3},−\mathrm{2},−\mathrm{1}\right\} \\ $$$$ \\ $$
Commented by MM42 last updated on 12/Apr/24
Commented by TheHoneyCat last updated on 14/Apr/24
Hey, sorry to be "mr bad news" here, but I think both your D and R are false...
i think you are missing a minus sign in D (no big deal, you probably just went too fast).
But more problematically, I believe your R is wrong. You must not remove "-3". This is visible on your own drawing, but also on the discussion Mr. W and I had previously (check f(-1/2) and you'll see).