How to study the critical points of a $2$-variable function?












0












$begingroup$


I am revising some past exam questions and there is one that states:




Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




According to my professor, this is what I have to do:




  1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


  2. Find the Hessian, using partial derivatives, which is
    begin{align*}
    f_{xx}(x,y) &= 2-12x^2-4y \
    f_{xy}(x,y) &= -8xy \
    f_{yx}(x,y) &= -8xy \
    f_{yy}(x,y) &= 2-12y^2-4x^2.
    end{align*}



And until here it's all good. However, what confuses me is the next step.



We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



Please explain in the clearest and simpelst way possible!



Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



Thanks, guys!






multivariable-calculus derivatives maxima-minima hessian-matrix







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am revising some past exam questions and there is one that states:




    Study the critical points of the function:
    $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




    According to my professor, this is what I have to do:




    1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


    2. Find the Hessian, using partial derivatives, which is
      begin{align*}
      f_{xx}(x,y) &= 2-12x^2-4y \
      f_{xy}(x,y) &= -8xy \
      f_{yx}(x,y) &= -8xy \
      f_{yy}(x,y) &= 2-12y^2-4x^2.
      end{align*}



    And until here it's all good. However, what confuses me is the next step.



    We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
    $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
    And here I'm lost. On the answer sheet it states:
    $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
    What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



    Please explain in the clearest and simpelst way possible!



    Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



    Thanks, guys!






    multivariable-calculus derivatives maxima-minima hessian-matrix







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am revising some past exam questions and there is one that states:




      Study the critical points of the function:
      $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




      According to my professor, this is what I have to do:




      1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


      2. Find the Hessian, using partial derivatives, which is
        begin{align*}
        f_{xx}(x,y) &= 2-12x^2-4y \
        f_{xy}(x,y) &= -8xy \
        f_{yx}(x,y) &= -8xy \
        f_{yy}(x,y) &= 2-12y^2-4x^2.
        end{align*}



      And until here it's all good. However, what confuses me is the next step.



      We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
      $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
      And here I'm lost. On the answer sheet it states:
      $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
      What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



      Please explain in the clearest and simpelst way possible!



      Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



      Thanks, guys!






      multivariable-calculus derivatives maxima-minima hessian-matrix







      share|cite|improve this question











      $endgroup$




      I am revising some past exam questions and there is one that states:




      Study the critical points of the function:
      $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




      According to my professor, this is what I have to do:




      1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


      2. Find the Hessian, using partial derivatives, which is
        begin{align*}
        f_{xx}(x,y) &= 2-12x^2-4y \
        f_{xy}(x,y) &= -8xy \
        f_{yx}(x,y) &= -8xy \
        f_{yy}(x,y) &= 2-12y^2-4x^2.
        end{align*}



      And until here it's all good. However, what confuses me is the next step.



      We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
      $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
      And here I'm lost. On the answer sheet it states:
      $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
      What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



      Please explain in the clearest and simpelst way possible!



      Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



      Thanks, guys!






      multivariable-calculus derivatives maxima-minima hessian-matrix




      multivariable-calculus derivatives maxima-minima hessian-matrix






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 14:05









      Rócherz

      2,7762721




      2,7762721










      asked Dec 8 '18 at 13:55









      BM97BM97

      758




      758






















          1 Answer
          1






          active

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          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43











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          1 Answer
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          1 Answer
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          active

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          active

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          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43
















          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43














          1












          1








          1





          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$



          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 13:59









          Siong Thye GohSiong Thye Goh

          101k1466117




          101k1466117












          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43


















          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43
















          $begingroup$
          Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:10




          $begingroup$
          Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:10












          $begingroup$
          Yes, they come from those places that you mentioned.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:12




          $begingroup$
          Yes, they come from those places that you mentioned.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:12












          $begingroup$
          Thank you !!! You are great!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:13




          $begingroup$
          Thank you !!! You are great!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:13












          $begingroup$
          May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
          $endgroup$
          – BM97
          Dec 8 '18 at 14:41




          $begingroup$
          May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
          $endgroup$
          – BM97
          Dec 8 '18 at 14:41












          $begingroup$
          The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:43




          $begingroup$
          The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:43


















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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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