Question Number 206328 by mnjuly1970 last updated on 12/Apr/24

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)
Commented by mr W last updated on 12/Apr/24

Commented by TheHoneyCat last updated on 12/Apr/24

Commented by mr W last updated on 12/Apr/24

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)
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).