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f-x-1-x-x-2-x-x-2-is-given-D-f-domain-R-f-range-




Question Number 206328 by mnjuly1970 last updated on 12/Apr/24
         f(x)= ⌊ (( 1+ x +x^2 )/(x+ x^( 2) )) ⌋ is given.         ⇒ { ((  D_f  = ? (domain ))),((   R_( f)  = ?( range))) :}
f(x)=1+x+x2x+x2isgiven.{Df=?(domain)Rf=?(range)
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}    _□ )) :}
D=RSoweneed1+x+x2x+x2=1x+x2+1RSoweneed1x×11+xRThisworksonallR{0,1}buywecouldalsoadd{+,}andalso{xCArg(x)=Arg(1+x)}Thisboilsdownto:xeiθ(1+xeiθ)Reiθ+xei2θRsinθ+xsin(2θ)=0Butassumingyoujustworkwithrealnumbers:Df=R{1,0}=],1[]1,0[]0,[Rf=R1x+x2+1=R1x+x2+1=(Rx+x2xR{1,0})1+1Now,x+x2isapolynomialofdeg2ithasitsminimureachedat12ieRx+x2=[14,+[so:Rx+x2xR{1,0}=[1/4,0[R+so:(Rx+x2xR{1,0})1=],4]R+Thus:Rf=(],4]+1)(R++1)=(R4+1)(Z++1)=∣],3][1,[{Df=R{1,0}Rf=Z{2,1,0}◻
Commented by mr W last updated on 12/Apr/24
for which x such that f(x)=−3?
forwhichxsuchthatf(x)=3?
Commented by TheHoneyCat last updated on 12/Apr/24
f(−1/2)=⌊(1/((−1/2)+(−1/2)^2 ))+1⌋                     =⌊(1/(−1/2+1/4))+1⌋                     =⌊(1/(−1/4))+1⌋                     =⌊−4+1⌋                     =⌊−3⌋                     =−3  But since we are talking about it,  notice that this point is very special.  Most points x are in a constant neigboorhood.  meaning x±ε with epsilon tiny enough   will give the same value.  some points are such that this holds only  for x+ε or x−ε . (points where the ⌊•⌋ causes  a jump)  But x_0 =−1/2 is the only point  that has a compleet neigboorhood that is worth  −4 and never itself.    this means lim_(x→x_0 ) f(x)=−4≠f(x_0 )    In more fancy words, f is continuous by part  everywhere except at x_0 .
f(1/2)=1(1/2)+(1/2)2+1=11/2+1/4+1=11/4+1=4+1=3=3Butsincewearetalkingaboutit,noticethatthispointisveryspecial.Mostpointsxareinaconstantneigboorhood.meaningx±ϵwithepsilontinyenoughwillgivethesamevalue.somepointsaresuchthatthisholdsonlyforx+ϵorxϵ.(pointswherethecausesajump)Butx0=1/2istheonlypointthathasacompleetneigboorhoodthatisworth4andneveritself.thismeanslimxx0f(x)=4f(x0)Inmorefancywords,fiscontinuousbyparteverywhereexceptatx0.
Commented by mr W last updated on 12/Apr/24
great!
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}
f(x)=[1x2+x]+1ifg(x)=1x2+xf(x)=[g(x)]+1Df=Dg=R{0,1}x{1g+1+g&x{0g0+g+g(x)=2x1(x2+x)2M(12,4)Rg=(,4](0,)Rf=Z{3,2,1}
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).

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