Critical points and extremum of $f(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2$

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Let $f:BbbR^2to BbbR$ be a function defined by
beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
$i.$ Compute the critical points of $f$



$ii.$ Does $f$ have an extremum?



My work:
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have
beginalignx=frac12(3x^2+2xy)endalign
Substituting into $2$, we get
beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign



I don't know where to go from here. Can someone please, help?







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    Let $f:BbbR^2to BbbR$ be a function defined by
    beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
    $i.$ Compute the critical points of $f$



    $ii.$ Does $f$ have an extremum?



    My work:
    beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
    beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
    At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
    we have
    beginalignx=frac12(3x^2+2xy)endalign
    Substituting into $2$, we get
    beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
    beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign



    I don't know where to go from here. Can someone please, help?







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $f:BbbR^2to BbbR$ be a function defined by
      beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
      $i.$ Compute the critical points of $f$



      $ii.$ Does $f$ have an extremum?



      My work:
      beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
      beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
      At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
      we have
      beginalignx=frac12(3x^2+2xy)endalign
      Substituting into $2$, we get
      beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
      beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign



      I don't know where to go from here. Can someone please, help?







      share|cite|improve this question














      Let $f:BbbR^2to BbbR$ be a function defined by
      beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
      $i.$ Compute the critical points of $f$



      $ii.$ Does $f$ have an extremum?



      My work:
      beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
      beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
      At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
      we have
      beginalignx=frac12(3x^2+2xy)endalign
      Substituting into $2$, we get
      beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
      beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign



      I don't know where to go from here. Can someone please, help?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 23 at 4:36

























      asked Aug 23 at 4:29









      Mike

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      75615




















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          beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
          beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
          At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
          we have (1) implies
          beginalignx(3x+2y-2)=0endalign
          So there are two cases $x=0$ or ... continue from there.






          share|cite|improve this answer




















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            beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
            beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
            At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
            we have (1) implies
            beginalignx(3x+2y-2)=0endalign
            So there are two cases $x=0$ or ... continue from there.






            share|cite|improve this answer
























              up vote
              3
              down vote













              beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
              beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
              At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
              we have (1) implies
              beginalignx(3x+2y-2)=0endalign
              So there are two cases $x=0$ or ... continue from there.






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
                beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
                At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
                we have (1) implies
                beginalignx(3x+2y-2)=0endalign
                So there are two cases $x=0$ or ... continue from there.






                share|cite|improve this answer












                beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
                beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
                At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
                we have (1) implies
                beginalignx(3x+2y-2)=0endalign
                So there are two cases $x=0$ or ... continue from there.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 23 at 4:39









                Andrew Allen

                1197




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