discriminant critical points

Show that (0;0) is a critical point. Not all critical points are local extrema. To make this process easier, let's make some substitutions. endobj Figure $\PageIndex{4}$: $z = x^2 + y^2$ has an absolute minimum of $0$ at $ (0,0)$, while $z = -(x^2 + y^2)$ has an absolute maximum of $0$ at $ (0,0)$, Example $\PageIndex{1}$: Classifying the critical points of a function. >> endobj /Filter /FlateDecode Donate or volunteer today! Since âf = hâ2x,â2yi, the only solution to âf = h0,0i is x = 0, /Type /Page b)If there is a local minimum, what is the value of the discriminant D at the point? >> Find the critical points by solving the simultaneous equations f y(x, y) = 0. 3. This formula is called the Second Partials Test, and it can be used to classify the behavior of any function at its critical points, as long as its second partials exist there and as long as the value of this discriminate is not zero. We can argue that it has an absolute minimum value of $-14$ at the point $ (3, -5) $, since we are adding squared terms to $-14$ and thus cannot get a value less than $-14$ for any values of $x$ and $y$, while we do obtain this minimum value of $-14$ at the vertex point $ (3, -5) $. c)If there is a local maximum, what is the value of the discriminiant D at that point? �8�*��I?b`J��H��+��SJ@ r8��\ s��r��`�w��G��2}�s� ��c��Ӊ�Sy:�U��JԫfZ��m帠�� 1L$�a��)�3/��V;��8t�Y®�7�G�!��dl�l��ψLրB��S]��v�� Here is a graph of the function. /Contents 28 0 R a) How many critical points does f have in R2? ? Calculate the discriminant for each critical point of Apply (Figure) to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive. d. Setting the partials of $f$ equal to $0$, we get: \[ \begin{align*} \text{Set}\quad f_x(x,y) &= 2x + 6y = 0 & \\ \text{and}\quad f_y(x,y) &= 6x + 2y = 0 & \implies y &= -3x \end{align*} \]Substituting $-3x$ into the first equation for $y$ gives us, \[\begin{align*}2x + 6(-3x) &= 0 \\ -16x &= 0 \\ x &= 0\end{align*}\]Since $y = -3x$, we have $ y = -3(0) = 0$, so the critical point of $f$ is $ (0,0) $. Classify each critical point with 2nd derivative. If the original function has a relative minimum at this point, so will the quadratic approximation, and if the original function has a saddle point at this point, so will the quadratic approximation. E��w�:|��ۙб�C��u��t'��J��+&��#�[��&�? if $a = \frac{f_{xx}(x_0, y_0)}{2} \lt 0$, the function $f$ opens downwards with a local maximum at the critical point $ (x_0, y_0) $. /Length3 532 *��%�ܟoK��T;�6~��.q��Zk^�s�d��̢��j�)L��##(蔪X�a��^�UWȨy':�5�\9��) �RE�[�H��[5�u��6��ek� j"SM�sf�@�Њ�#މDF��r�Dɗ��E�;+��j��8�4�pP��s�]Z�!��s�R��Ѹ��2�º�[\y��Ɏ��B�I[6�k�w��9B(�d5�NR��7�;5��R؆�0�QR�bN��蔓tu]��v��{I�G�ix��$M�h�k�B 2 0 obj << If you're seeing this message, it means we're having trouble loading external resources on our website. Since sharing the same second partials means the two surfaces will share the same concavity (or curvature) at the critical point, this causes these quadratic approximation surfaces to share the same behavior as the function $z = f(x, y)$ that they approximate at the point of tangency. �x��%��iS)~M�� iH�g��T�M�kLCps��A�R�!S�� (��T��}�� ,9V��M�ıb2=Q�-fz�n��교*�x{��J��=��(`#;��A�͔�)��jbF$��3�J��Hl � �QD� �ZPĥ*R K� b�\A�ހ�"��l��g��w��Zs��x�:4��IG�y��(��m��I��a��d �� 'Ùf�\�%H``?Er~"T�!0�aR��( �h�PRN~B!��:!Q��. The number of discriminant The critical points of this function of yare found by setting the derivative to zero: @ @y (3+2y2 4y) = 0 =)4y 4 = 0 =)y= 1 with f( 1;1) = 1 : the line x= 1: f(1;y) = 2y2 1: Computing the derivative and setting it to 0 we ï¬nd the critical point y= 0. ��>Ţ��#Ǘ��bQ��u�x��!6)�Wf�RPLS]� �h�ᘭ��!�� V�t�#��A�Z#$��j��;QI��ܠB�5Q8��'�� :�W�1@:4X_��1A{�@��S�fF�Y�Eb��;,��aA�}�tM��ɉ�9��C��%많��l$��wT�(J?p̏��y��T��-��^��[�)5��}L|��Fq9A^��8�ƺ��eW�ۥ9��.��5UW�\v��|=}Q�D��* �Odx�]�C�%��˕��}4�f��ʈ��ߡ�zI�l��2�jJ�7D,u)�i�Mp�`xơ�g�@�A��6@��rgg��"T-��P»o �Cr~d)�B�5��Z \[\begin{align*} f(x,y) &= x^2 + 6xy + y^2 \\ &= (x^2 + 6xy + 9y^2) + y^2 - 9y^2 \\ &= (x + 3y)^2 - 8y^2 \end{align*}\]As this produces a difference of squares with one positive squared term and the other a negative squared term, we see that $f$ takes a form similar to $z = x^2 - y^2$ and will have a saddle point at $ (0, 0, 0) $. To complete the square here, we first need to factor out the factors of the squared terms. One of the critical points of the function is (0,0) and the discriminant is zero. }�N�F�v`�i5��e��;��n�>{*M`N�1vu�Ҝ3 �C��ܭ//�^%��T�A�P� ��|0�6z"_m��0��-��l��&��w_ⴸM[ ߈VM;�9�#2�9�b� �O̠%t� &�l�ױ4�$yt��$;��{�Kh�0p�Oir`�rW�d��d��9o��;�{�"�f��1 x�]��.ӷG��-�[Ŧ��Nҫ�7}{�3��E�1㥥ݍb��%vi��ْ��m��uφ�vG��}�/��3o�R�8��K��2C.�Z[�Ԧ l�E�v�� K�:e2Y�v�rO��O�J�ˢ�T6[-��NW_�3rX�+��t6D)�v��fʺ��t�0��'�#�=��x5��'�C�͖��*6O�ۦ��jMf�Ĳˠ�W�Cb�a��5��R�y�R�� Wyp�r��L��j��>�V�P$S �޾qw�-�硯E��?I��Eʝw�4��v��z��m�asJڍJհ�J3�-��50�}�r�e\'��\�`u�L��Z_`��k*�6@m��E�С��R�n��7�I�Z�\'磂��+�܄��$�b�s)��:�:�8�m�'�m�z��/S&��fws�͗��b?S�%SDY��&vjxy�m�^z�*�fr-�U�K7��u�uP�n�Y�Q��jx��!�p6{Z� ��IT��endstream But, since the point $ (x_0, y_0) $, in this case, is a critical point of $f$, we know that $f_x(x_0, y_0) = 0$ and $f_y(x_0, y_0) = 0$. x}O��O�۔�5�{ z�@�ޤ��V�cN}La� =ކ��w��^ǣ�d2R&�j�@VA3'L��r2��o�_��y��2�~�r��qT\jy��Wqd03�R��-H��Pi��H��f�ui B��Ғ��7��T��~kv�B��H��$��?�Xf��0e 1G#4&�:X��9\cnN�;BJ��#t��4�� 5Kw�h&��t[�G�"�2aND��V��-S�?TA�J��7f3iVG�K��oh�Ҥ�Jꙶ��-��n;����;3AJ1riت��Q��jo �P�� }Z�b�EV��Ȇs�L~��M�*L��5��E�m�#�լ��\}RĤ�D�cԸ��=��J�Uj��sE��Nm! In the last slide we saw that. if $a = \frac{f_{xx}(x_0, y_0)}{2} \gt 0$, the function $f$ opens upwards with a local minimum at the critical point $ (x_0, y_0) $. /Parent 25 0 R Find the maxima, minima and saddle points of z= (x2 y2)e( 2x2 y )=2. Exercise. If â(x 0,y Now there are really three basic behaviors of a quadratic polynomial in two variables at a point where it has a critical point. The difference of two squared terms, like $z = f(x, y) = x^2 - y^2$ or $z = f(x, y) = y^2 - x^2$, producing a saddle with a saddle point at its critical point. If there is none, type N. Answer: N Therefore, all critical points are saddle. Setting these equal to zero gives a system of equations that must be solved to find the critical points: y^2-6x+2=0, 2y(x-1)=0. How to determine if the critical point of a two-variable function is a local minimum, a local maximum, or a saddle point. In fact it is the specialization to a simple case of the general notion of critical point given below. Critical point of a single variable function. C�.�RDڃHRBHر�Wʒ�X��q2!�LH&@Ppå Click here to let us know! \[ \begin{align*} \text{Set}\quad f_x(x,y) &= 2x -6 = 0 & \implies x &= 3 \\ \text{and}\quad f_y(x,y) &= 2y + 10 = 0 & \implies y &= -5 \end{align*} \]We obtain a single critical point with coordinates $ (3, -5) $. Setting the partials of $f$ equal to $0$, we obtain: \[ \begin{align*} \text{Set}\quad f_x(x,y) &= -6x -6 = 0 & \implies x &= -1 \\ \text{and}\quad f_y(x,y) &= -2y + 12 = 0 & \implies y &= 6 \end{align*} \]We obtain a single critical point with coordinates $ (-1, 6) $. Completing the square, we get: \[\begin{align*} f(x,y) &= x^2 - 6x + y^2 + 10y + 20 \\ &= x^2 - 6x + 9 + y^2 + 10y + 25 + 20 - 9 - 25 \\ &= (x - 3)^2 + (y + 5)^2 - 14 \end{align*}\]Notice that this function is really just a translated version of $z = x^2 + y^2$, so it is a paraboloid that opens up with its vertex (minimum point) at the critical point $ (3, -5) $. /Font << /F17 6 0 R /F18 9 0 R /F15 12 0 R /F21 15 0 R /F22 18 0 R /F24 21 0 R /F19 24 0 R >> Determine the critical points of the function where Discard any points where at least one of the partial derivatives does not exist. Factoring out $-2$ from the $y$-squared term gives us: \[\begin{align*} f(x,y) &= x^2 + 8x - 2y^2 + 16y \\ &= x^2 + 8x +16 - 2\left(y^2 - 8y + 16\right) - 16 + 32 \\ &= (x + 4)^2 - 2(y - 4)^2 +16\end{align*}\]Since one squared term is positive and one is negative, we see that this function has the form of $z = x^2 - y^2$ and so it has a saddle point at its critical point. About. e) What is the maximum value of f on R2? All local extrema are critical points. To test such a point to see if it is a local maximum or minimum pointâ¦ %CRITCALPOINTS (f) is a function to determine the critical points of a 2D %surface given the function f (x,y). The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. \[Q(x, y) = f (x_0, y_0) + f_x(x_0, y_0) (x - x_0) + f_y(x_0, y_0) (y - y_0) + \frac{f_{xx}(x_0, y_0)}{2}(x-x_0)^2 + f_{xy}(x_0, y_0)(x-x_0)(y-y_0) + \frac{f_{yy}(x_0, y_0)}{2}(y-y_0)^2\]. If there is none, type N. (C) If there is a local maximum, what is the value of the discriminant D at that point? /Length1 757 76̙j��ra�H,"�sb�k2E�-�͘�[�a3NBf��Ed��ebOF0)�s9��8Ws��cb>�r�{�8��=` P=_ܢnAf��/��/Ag�_��u��_2�~��ܛw̌:Y��H�j�AO+:h��*�3n��k?��r��h�N�m��g+M3B3��g/A�٧��T_C�02��N�'�%��u#q�IO�u��c��x�� But if $D = 4ac-b^2 = 0$, the quadratic polynomial reduces to $Q(x,y) = a\left(u+ \frac{b}{2a}v\right)^2 + d$, whose graph is a parabolic cylinder, so the behavior of the function is not clear at the critical point $ (x_0, y_0) $. /MediaBox [0 0 595.276 841.89] #1. *Ar��2i�!�&S ��{�� Zơ�Kb5ˌ)�B�X��]�� ֋Ae��M Now let's consider the quadratic approximation to a function $z = f(x, y)$ centered at a critical point $ (x_0, y_0) $ of this function. D�%�X��K�sf�x=��J�=��ZI��Ť��7 �iE�E�Xؒ�d��xϠrK�a��E40��Dj�Kϕ ��%� 4#ӹ�=��5��Q��)�Ҭ��ڜ։ui�$)�~I-��h��G�L��Z��f�s��R�*�f��"��yP@�I�@��6�N+z��%�S��W��m�%1ˇ�~��~>�#��e�5E�ۓXypg��[��B�J��&��9��[�^�nWK��7X�;�+�H8�B��h�@�� w-�\��j�+@tK�1a��kq�E�^��re�0d]��=��lJ��S]_y��-2o�^w�H��ǻ�u��Y�Yn��r'k��N!�.�b�|��y�_�W�g�~Mzu��O�A�6h You get these gems as you gain rep from other members for making good contributions and giving helpful advice. Now we need to complete the square on this quadratic polynomial in two variables to learn how we can classify the behavior of this function at this critical point. 3 0 obj << In order to develop a general method for classifying the behavior of a function of two variables at its critical points, we need to begin by classifying the behavior of quadratic polynomial functions of two variables at their critical points. The only point that will make both of these derivatives zero at the same time is (0,0) ( 0, 0) and so (0,0) ( 0, 0) is a critical point for the function. ! The two first order partial derivatives are, f x(x,y) = y f y(x,y) = x f x ( x, y) = y f y ( x, y) = x. Up Next. â âyf(x,y)|x=a,y=b =0 â â y f ( x, y) | x = a, y = b = 0. ð. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Legal. If the experimental F exceeds a critical F, then the experimental groups can be distinguished based on the predictor variables. Note it would be similar to the form, $z = x^2 + y^2$. value of second partial derivative with respect to x or y. characterization Find and classify critical points Useful facts: The discriminant â = f xxf yy â f xy 2 at a critical point P(x 0,y 0) plays the following role: 1. Occurrence of local extrema: All local extrema occur at critical points, but not all critical points occur at local extrema. Site Navigation. D(x, y) what is the value of the discriminant at the critical point? f(x,y) = 3xy 2 +x 3-3x 2-3y 2 +2. The derivative at the critical point u = 3 is zero (as might be expected). please help (C) If there is a local maximum, what is the value of the discriminant D at that point? (c) Sketch contours near the critical point to determine, or confirm, whether it is a local maximum, a local minimum, a saddle point, or none of these. Not all critical points give rise to local minima/maxima De nition: Saddle Point A di erentiable function f(x;y) has a saddle point at a critical point (a;b) if in every open disk ... 2 is called the discriminant or Hessian of f. It is sometimes easier to remember it in determinant form, f xxf yy f xy 2 = f xx f xy f xy f Reasoning behind second partial derivative test. /Length2 1104 Example 13.7.2: Using the Second Derivative Test Use completing the square to identify local extrema or saddle points of the following quadratic polynomial functions: a. (x, y) = (b) Find the discriminant, D(x, y) for the function. T�rKg#^g*J@ /Length 1668 value of discriminant. >> endobj See Figure $\PageIndex{3}$. x��RiX�. d) If there is a saddle point, what is the value of the discriminant D at that point. A critical point $x = c$ is an inflection point if the function changes concavity at that point. /ProcSet [ /PDF /Text ] When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. /Contents 3 0 R The corre-sponding point (1;0) is one of the corners, and we will consider it separately below. It will fit one of the following three forms, often being a transformation of one of the following functions. Note it would be similar to the form, $z = -\left(x^2 + y^2\right)$. Next we need to determine the behavior of the function $f$ at this point. Discriminant function analysis is used to determine which continuous variables ... equal the number of data points in all groups minus the number of groups (N â k). The point (a,b) ( a, b) is a critical point for the multivariable function f(x,y), f ( x, y), if both partial derivatives are 0 at the same time. To determine the behavior of $f$ at this critical point, we complete the square. b��*�~4�NTtN��_�� j��r&�\h*J,��.=��c9 ��@�؏IQ��eCOQ�}�\��i��b��C�܄�L ��z�l��exR��3��R��Ef G(�J��n�`��P5�~��~��վB��v�꾦B!�n�o��ڰ�9n��-�K�.��V��a|+}� ��NAy��v_�8Ξ��,u~P�V"oy�L��:��[��Kx��$�O�;��i�\�c�Q �Kw�HI{:y�l��p܈�:��a^s0�F��5"mo��,�ɸ��#V�!/�V7 ��ۦϔ8i��ƨ:2�kʦt��^��]0�Z�4�0 News; ,�#��9�R/[� This expression, $4ac-b^2$, is called the discriminant, as it helps us discriminate (tell the difference between) which behavior the function has at this critical point. ?R�x�W��{��]�dg��N�/�L��6�.+�Z?�x��'�zl�� =�SΨ'%��sI��K��;2B��qB�.�_ɠYdP�r��zŖo_��DoaUR�I�e�R��z*�A�i,�|>d�J|о�_Q��x��-��1�� m��b�%Eȱ�c��*�� !^D�}!=��A6��0�/��QU��>٣ؿS�m��y��HH�T��#T��&o}/�3��>�"�c��^��WtB�hnyCWӎG��j^$*%>}Np ��5�k1N��v��-��ӨE�L�N��Y ��]}�M��-��Ή��#��W� [V �z�7?��bendstream The quadratic equation has no roots as the discriminant $D = 16 â 20 = â 4 \lt 0.$ Note that $x = 2$ is a not a critical point as the function is not defined at this point. Reasoning behind second partial derivative test. /Type /Page #&�V��$��Y��q-��o�<3�� qd�b��ܑ�C�su�}�jPx�И��!��W1�}9a�� ;�ɤ�5z~Ca��4"v��㊫��U�Cz3�^cbsId6�� M�-��[͖�I��{J�v�s��FT nֻj��C/�*�ѽ�~��rw&��1>��YE �CHάu��O��6]k��j�/i��Gk_?m�02>εsɜ�߽�7v�`L��Un�\W�V�V��GF�P=�GTOs?�{�y:�� ֘Y ��l��! /Font << /F21 15 0 R /F15 12 0 R /F19 24 0 R /F24 21 0 R /F22 18 0 R /F18 9 0 R /F25 31 0 R /F20 34 0 R >> Find local maxima and minima and saddle points of the function. Have questions or comments? A sum of two squared terms, like $z = x^2 + y^2$, producing a paraboloid that opens up and has a relative (absolute) minimum at its vertex. If â(x 0,y 0) > 0 and f xx(x 0,y 0) > 0, then f has a local minimum at (x 0,y 0). This means that even if the surface is concave up in both $x$- and $y$-directions, or concave down in both $x$- and $y$-directions, a large mixed partial can offset these and cause the surface to have a saddle point at the point $(x_0, y_0)$. Critical points play an important role in the study of plane curves defined by implicit equations, in particular for sketching them and determining their topology. #1. The determinant of the Hessian at x is called, in some contexts, a discriminant. Remember that the original function will share the same behavior (max, min, saddle point) as this 2nd-degree Taylor polynomial at this critical point. x��\Yo�~��#/;}>Hk y��cE@��>U�s��T�E��sL��Q]�W_W[,8��p�mX\ܽ��? Find the critical points of the function f (x,y) = âx2 â y2. (B) If there is a local minimum, what is the value of the discriminant D at that point? >> In order to perform the classification efficiently, we create inlinevectorized versions of the Hessian determinant and of the second partial In other words, if the original function has a relative maximum at this point, so will the quadratic approximation. To determine the critical points of this function, we start by setting the partials of $f$ equal to $0$. /MediaBox [0 0 595.276 841.89] We can argue that it has an absolute maximum value of $51$ at the point $ (-1, 6) $, since we are subtracting squared terms from $51$ and thus cannot get a value more than $51$ for any values of $x$ and $y$, while we do obtain this minimum value of $51$ at the vertex point $ (-1, 6) $. The discriminant for any quadratic equation of the form $$ y =\red a x^2 + \blue bx + \color {green} c $$ is found by the following formula and it provides critical information regarding the nature of the roots/solutions of any quadratic equation. >> endobj Critical points If the gradient (the vector of the partial derivatives) of a function f is zero at some point x, then f has a critical point (or stationary point) at x. ?z�/M�y$�Ԧ�8Sw�a[�y\Ʉ|�օ��#�MO웙P-}:��)\��r�S�*]M�XX��(8��m�S|8�x�N�0z�"�#�=��3.Y��)�̑̅0_b#̑bNb�Sb�́V��+*WS�:�2%�!nY-��B��`��!4g�N�^��0O2�!��s��@��+��VeFN�� 0ǋ�؍&Q�z�1��\��z��4�q��"��E��tMjZ��*�Ψ֥=�\ƯeQ��g�y=�_>��/mi? Critical Points: The extremes of a function can be maximum or minimum. Adopted a LibreTexts for your class? Rewriting the perfect square trinomial as the square of a binomial and combining the $v^2$ terms yields: \[\begin{align*} &= a\left[ \left(u+ \frac{b}{2a}v\right)^2 + \left(\frac{c}{a} - \frac{b^2}{4a^2}\right)v^2 \right] + d \\ &= a\left[ \left(u+ \frac{b}{2a}v\right)^2 + \left(\frac{4ac}{4a^2} - \frac{b^2}{4a^2}\right)v^2 \right] + d \\ And the sign of this coefficient is determined only by its numerator, as the denominator is always positive (being a perfect square). %The method choosen is to compute the first and second partial derivatives %on the given function by first evaluating the Jacobian and Hessian Matrix Since a critical point (x0,y0) is a solution to both equations, both partial derivatives are zero there, so that the tangent plane to the graph of f(x, y) is horizontal. Exercise. A critical point or stationary point of a differentiable function of a single real variable, f(x), is a value x 0 in the domain of f where its derivative is 0: f â²(x 0) = 0.A critical value is the image under f of a critical point. Maxima, minima, and saddle points. How to determine if the critical point of a two-variable function is a local minimum, a local maximum, or a saddle point. /Filter /FlateDecode If â(x 0,y 0) > 0 and f xx(x 0,y 0) < 0, then f has a local maximum at (x 0,y 0). Example 6.3.1. See the plot on the right side of Figure $\PageIndex{4}$. (1 point) Suppose f(x, y) = x2 + y2 â 8x â 2y + 3 (A) How many critical points does f have in R2? Next we need to determine the behavior of the function $f$ at this point. Critical points are places where $\nabla f = 0$ or $\nabla f$ does not exist. Calculate the discriminant D = fxx(x0, y0)fyy(x0, y0) â (fxy(x0, y0))2 for each critical point of f. Apply the four cases of the test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive. 26 0 obj << 2. value of the function. If D = 4ac â b2 > 0, then the two squared terms inside the brackets are both positive, and For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. That is, $f$ has a saddle point at $ (-4, 4, 16) $. /Resources 26 0 R In either case, the quadratic polynomial will be in the form of $z = x^2 - y^2$ or $z = y^2 - x^2$ (i.e., it will be the difference of two squared terms), so we get a saddle point at the critical point $ (x_0, y_0) $. 1. >> endobj Examples for use of Lagrangeâs multiplier method. endobj /Length 3251 33 0 obj << See the plot on the left side of Figure $\PageIndex{4}$. The notion of critical point that is used in this section, may seem different from that of previous section. Let's choose to let $u = x - x_0$ and $v = y - y_0$, and let \[\begin{align*} a &= \frac{f_{xx}(x_0, y_0)}{2}, \\ b &= f_{xy}(x_0, y_0), \\ c &= \frac{f_{yy}(x_0, y_0)}{2} \,\text{and} \\ d &= f (x_0, y_0) \end{align*}\], Then we need to complete the square on the polynomial: \[ Q(x,y) = au^2 +buv + cv^2 + d\], First we factor out the coefficient of $u^2$: \[= a\left[ u^2 + \frac{b}{a}uv + \frac{c}{a}v^2\right] + d\], Next, we complete the square using the first two terms: \[= a\left[ \left(u^2 + \frac{b}{a}uv + \left(\frac{b}{2a}v\right)^2\right) + \frac{c}{a}v^2 - \left(\frac{b}{2a}v\right)^2 \right] + d\]. ?. Critical points are where the tangent plane to $z=f (x,y)$ is horizontal or does not exist.
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