{"id":3936,"date":"2022-04-05T18:52:48","date_gmt":"2022-04-05T18:52:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3936"},"modified":"2022-10-29T02:15:19","modified_gmt":"2022-10-29T02:15:19","slug":"second-derivative-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/second-derivative-test\/","title":{"raw":"Second Derivative Test","rendered":"Second Derivative Test"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167794099268\">Consider the function\u00a0[latex]f(x)=x^{3}[\/latex].\u00a0This function has a critical point at [latex]x=0[\/latex], since [latex]f'(0)=3(0)^{3}=0[\/latex]. However, [latex]f[\/latex] does not have an extreme value at [latex]x=0[\/latex]. Therefore, the existence of a critical value at [latex]x=x_0[\/latex] does not guarantee a local extremum at [latex]x=x_0[\/latex]. The same is true for a function of two or more variables. One way this can happen is at a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term197\" data-type=\"term\">saddle point<\/span>.<\/strong> An example of a saddle point appears in the following figure.<\/p>\r\n\r\n[caption id=\"attachment_1279\" align=\"aligncenter\" width=\"483\"]<img class=\"size-full wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134416\/4-7-3.jpeg\" alt=\"x^{2}-y^{2}\" width=\"483\" height=\"651\" \/> Figure 1. Graph of the function\u00a0[latex]\\small{z=x^{2}-y^{2}}[\/latex].\u00a0This graph has a saddle point at the origin.[\/caption]In this graph, the origin is a saddle point. This is because the first partial derivatives of [latex]f(x, y)=x^{2}-y^{2}[\/latex] are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to [latex]y=0[\/latex] is [latex]z=x^{2}[\/latex] (a parabola opening upward), but the vertical trace corresponding to [latex]x=0[\/latex] is [latex]z=-y^{2}[\/latex] (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nGiven the function [latex]z=f(x, y)[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>the point [latex](x_0, y_0, f(x_0, y_0))[\/latex] is a saddle point if both [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>but [latex]f[\/latex] does not have a local extremum at [latex](x_0, y_0)[\/latex].\r\n\r\n<\/div>\r\nThe second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function.\u00a0 Below we recall the the second derivative test as it applies to single-variable functions to note the similarities to its two-variable extension.\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\">Recall: Second Derivative Test (Single Variable Version)<\/h3>\r\nSuppose that [latex] f(x) [\/latex] is twice-differentiable at [latex] x = x_0 [\/latex] and [latex] f'(x_0) = 0 [\/latex].\r\n<ol>\r\n \t<li>If [latex] f''(x_0) &lt; 0 [\/latex], then [latex]f [\/latex] has a local maximum at [latex] x_0 [\/latex].<\/li>\r\n \t<li>If [latex] f''(x_0) &gt; 0 [\/latex], then [latex]f [\/latex] has a local minimum at [latex] x_0 [\/latex].<\/li>\r\n \t<li>If [latex] f''(x_0) = 0 [\/latex], then the test is inconclusive.<\/li>\r\n<\/ol>\r\n<\/div>\r\nWhen extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term198\" data-type=\"term\">discriminant [latex]D[\/latex]\u00a0<\/span><\/strong>that replaces [latex]f''(x_0)[\/latex] in the second derivative test for a function of one variable.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: Second derivative test<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]. Define the quantity\r\n<p style=\"text-align: center;\">[latex]\\large{D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2}.[\/latex]<\/p>\r\n\r\n<ol id=\"fs-id1167793417260\" type=\"i\">\r\n \t<li>If\u00a0[latex]D&gt;0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)&gt;0[\/latex], then\u00a0[latex]f[\/latex] has a local minimum at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D&gt;0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)&lt;0[\/latex], then\u00a0[latex]f[\/latex] has a local maximum at\u00a0[latex](x_0,y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D&lt;0[\/latex], then\u00a0[latex]f[\/latex] has a saddle point at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D=0[\/latex], then the test is inconclusive.<\/li>\r\n<\/ol>\r\nSee\u00a0Figure 2.\r\n\r\n<\/div>\r\n\r\n[caption id=\"attachment_1281\" align=\"aligncenter\" width=\"943\"]<img class=\"size-full wp-image-1281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134549\/4-7-4.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. Figure a has two bulbous mounds pointing down, and the two extrema are listed as the local minima. Figure b has two bulbous mounds pointed up, and the two extrema are listed as the local maxima. Figure c is shaped like a saddle, and in the middle of the saddle, a point is marked as the saddle point.\" width=\"943\" height=\"345\" \/> Figure 2. The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).[\/caption]\r\n\r\nTo apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.\r\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\">Problem-solving strategy: using the second derivative test for functions of two variables<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet\u00a0[latex]z=f(x,y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. To apply the second derivative test to find local extrema, use the following steps:\r\n<ol>\r\n \t<li>Determine the critical points [latex](x_0, y_0)[\/latex]\u00a0of the function [latex]f[\/latex] where [latex]f_x(x_0, y_0)=f_y(x_0, y_0)[\/latex]. Discard any points where at least one of the partial derivatives does not exist.<\/li>\r\n \t<li>Calculate the discriminant [latex]D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2[\/latex]\u00a0for each critical point of [latex]f[\/latex].<\/li>\r\n \t<li>Apply\u00a0Second Derivative Test\u00a0to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using the second derivative test<\/h3>\r\n<p id=\"fs-id1167793244016\">Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:<\/p>\r\n\r\n<ol id=\"fs-id1167793374827\" type=\"a\">\r\n \t<li>[latex]f(x, y)=4x^{2}+9y^{2}+8x-36y+24[\/latex]<\/li>\r\n \t<li>[latex]g(x, y)=\\frac{1}{3}x^{3}+y^{2}+2xy-6x-3y+4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"914523760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"914523760\"]\r\n\r\na. Step 1 of the problem-solving strategy involves finding the critical points of [latex]f[\/latex]. To do this, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set each of them equal to zero:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\nf_x(x,y)&amp;=8x+8 \\\\\r\n\r\nf_y(x,y)&amp;=18y-36.\r\n\r\n\\end{align}[\/latex]\r\n\r\nSetting them equal to zero yields the system of equations\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n8x+8&amp;=0 \\\\\r\n\r\n18y-36&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThe solution to this system is [latex]x=-1[\/latex] and [latex]y=2[\/latex]. Therefore [latex](-1, 2)[\/latex] is a critical point of [latex]f[\/latex].<span data-type=\"newline\">\r\n<\/span>Step 2 of the problem-solving strategy involves calculating [latex]D[\/latex]. To do this, we first calculate the second partial derivatives of [latex]f[\/latex]:<span data-type=\"newline\">\r\n<\/span>\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\nf_{xx}(x,y)&amp;=8 \\\\\r\n\r\nf_{xy}(x,y)&amp;=0 \\\\\r\n\r\nf_{yy}(x,y)&amp;=18.\r\n\r\n\\end{align}[\/latex]\r\n\r\nTherefore, [latex]D=f_{xx}(-1,2)f_{yy}(-1,2)-(f_{xy}(-1,2))^2=(8)(18)-(0)^2=144[\/latex].\r\n\r\nStep 3 states to check the\u00a0Second Derivative Test for Functions of Two Variables.. Since [latex]D&gt;0[\/latex] and [latex]f_{xx}(-1,2)&gt;0[\/latex], this corresponds to case 1. Therefore, [latex]f[\/latex] has a local minimum at [latex](-1, 2)[\/latex] as shown in the following figure.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1283\" align=\"aligncenter\" width=\"813\"]<img class=\"wp-image-1283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135113\/4-7-5.jpeg\" alt=\"The function f(x, y) = 4x2 + 9y2 + 8x \u2013 36y + 24 is shown with local minimum at (\u20131, 2, \u201316). The shape is a plane curving up on both ends parallel to the y axis.\" width=\"813\" height=\"405\" \/> Figure 3. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum at [latex]\\small{(-1,2,-16)}[\/latex].[\/caption]b. For step 1, we first calculate [latex]g_x(x,y)[\/latex] and [latex]g_y(x,y)[\/latex], then set each of them equal to zero:[latex]\\hspace{8cm}\\begin{align}g_x(x,y)&amp;=x^2+2y-6 \\\\g_y(x,y)&amp;=2y+2x-3.\\end{align}[\/latex]\r\nSetting them equal to zero yields the system of equations\r\n[latex]\\hspace{8cm}\\begin{align}x^2+2y-6&amp;=0 \\\\2y+2x-3&amp;=0.\\end{align}[\/latex]\r\nTo solve this system, first solve the second equation for [latex]y[\/latex]. This gives [latex]y=\\frac{3-2x}{2}[\/latex]. Substituting this into the first equation gives[latex]\\hspace{7.5cm}\\begin{align}x^2+3-2x-6&amp;=0 \\\\x^2-2x-3&amp;=0 \\\\(x-3)(x+1)&amp;=0.\\end{align}[\/latex]\r\nTherefore, [latex]x=-1[\/latex] or [latex]x=3[\/latex]. Substituting these values into the equation\u00a0[latex]y=\\frac{3-2x}{2}[\/latex] yields the critical points [latex](-1,\\frac52)[\/latex] and [latex](3,-\\frac32)[\/latex].\r\n<p id=\"fs-id1167794138811\">Step 2 involves calculating the second partial derivatives of [latex]g[\/latex]:<\/p>\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\ng_{xx}(x,y)&amp;=2x \\\\\r\n\r\ng_{xy}(x,y)&amp;=2 \\\\\r\n\r\ng_{yy}(x,y)&amp;=2.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThen, we find a general formula for\u00a0[latex]D[\/latex]:\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\nD&amp;=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)-(g_{xy}(x_0,y_0))^2 \\\\\r\n\r\n&amp;=(2x_0)(2)-2^2 \\\\\r\n\r\n&amp;=4x_0-4.\r\n\r\n\\end{align}[\/latex]\r\n\r\nNext, we substitute each critical point into this formula:\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\nD(-1,\\frac52)&amp;=(2(-1))(2)-(2)^2=-4-4=-8 \\\\\r\n\r\nD(3,-\\frac32)&amp;=(2(3))(2)-(2)^2=12-4=8.\r\n\r\n\\end{align}[\/latex]\r\n\r\nIn step 3, we note that, applying the Second Derivative Test for Functions of Two Variables to point [latex](-1,\\frac{5}{2})[\/latex] leads to case 3, which means that [latex](-1,\\frac{5}{2})[\/latex]\u00a0is a saddle point. Applying the theorem to point [latex](3,-\\frac{3}{2})[\/latex]\u00a0leads to case 1, which means that [latex](3,-\\frac{3}{2})[\/latex] corresponds to a local minimum as shown in the following figure.\r\n\r\n[caption id=\"attachment_1284\" align=\"aligncenter\" width=\"829\"]<img class=\"size-full wp-image-1284\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135248\/4-7-6.jpeg\" alt=\"The function f(x, y) = (1\/3)x3 + y2 + + 2xy \u2013 6x \u2013 3y + 4 is shown with local minimum at (3, \u20133\/2, \u201329\/4) and saddle point at (\u22121, 5\/2, 41\/12). The shape is a plane curving up on the corners near (4, 3) and (\u22122, \u22122).\" width=\"829\" height=\"488\" \/> Figure 4. The function [latex]\\small{g(x,y)}[\/latex] has a local minimum and a saddle point.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nUse the second derivative to find the local extrema of the function\r\n\r\n[latex]f(x, y)=x^{3}+2xy-6x-4y^{2}[\/latex].\r\n\r\n[reveal-answer q=\"233578991\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233578991\"]\r\n\r\n[latex](\\frac{4}{3},\\frac{1}{3})[\/latex] is a saddle point,\u00a0[latex](-\\frac{3}{2},-\\frac{3}{8})[\/latex] is a local maximum.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186168&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0IESE6fWxpQ&amp;video_target=tpm-plugin-0hltm37e-0IESE6fWxpQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.35_transcript.html\">transcript for \u201cCP 4.35\u201d here (opens in new window).<\/a><\/center><\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167794099268\">Consider the function\u00a0[latex]f(x)=x^{3}[\/latex].\u00a0This function has a critical point at [latex]x=0[\/latex], since [latex]f'(0)=3(0)^{3}=0[\/latex]. However, [latex]f[\/latex] does not have an extreme value at [latex]x=0[\/latex]. Therefore, the existence of a critical value at [latex]x=x_0[\/latex] does not guarantee a local extremum at [latex]x=x_0[\/latex]. The same is true for a function of two or more variables. One way this can happen is at a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term197\" data-type=\"term\">saddle point<\/span>.<\/strong> An example of a saddle point appears in the following figure.<\/p>\n<div id=\"attachment_1279\" style=\"width: 493px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1279\" class=\"size-full wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134416\/4-7-3.jpeg\" alt=\"x^{2}-y^{2}\" width=\"483\" height=\"651\" \/><\/p>\n<p id=\"caption-attachment-1279\" class=\"wp-caption-text\">Figure 1. Graph of the function\u00a0[latex]\\small{z=x^{2}-y^{2}}[\/latex].\u00a0This graph has a saddle point at the origin.<\/p>\n<\/div>\n<p>In this graph, the origin is a saddle point. This is because the first partial derivatives of [latex]f(x, y)=x^{2}-y^{2}[\/latex] are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to [latex]y=0[\/latex] is [latex]z=x^{2}[\/latex] (a parabola opening upward), but the vertical trace corresponding to [latex]x=0[\/latex] is [latex]z=-y^{2}[\/latex] (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>Given the function [latex]z=f(x, y)[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>the point [latex](x_0, y_0, f(x_0, y_0))[\/latex] is a saddle point if both [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>but [latex]f[\/latex] does not have a local extremum at [latex](x_0, y_0)[\/latex].<\/p>\n<\/div>\n<p>The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function.\u00a0 Below we recall the the second derivative test as it applies to single-variable functions to note the similarities to its two-variable extension.<\/p>\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\">Recall: Second Derivative Test (Single Variable Version)<\/h3>\n<p>Suppose that [latex]f(x)[\/latex] is twice-differentiable at [latex]x = x_0[\/latex] and [latex]f'(x_0) = 0[\/latex].<\/p>\n<ol>\n<li>If [latex]f''(x_0) < 0[\/latex], then [latex]f[\/latex] has a local maximum at [latex]x_0[\/latex].<\/li>\n<li>If [latex]f''(x_0) > 0[\/latex], then [latex]f[\/latex] has a local minimum at [latex]x_0[\/latex].<\/li>\n<li>If [latex]f''(x_0) = 0[\/latex], then the test is inconclusive.<\/li>\n<\/ol>\n<\/div>\n<p>When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term198\" data-type=\"term\">discriminant [latex]D[\/latex]\u00a0<\/span><\/strong>that replaces [latex]f''(x_0)[\/latex] in the second derivative test for a function of one variable.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: Second derivative test<\/h3>\n<hr \/>\n<p>Let [latex]z=f(x, y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]. Define the quantity<\/p>\n<p style=\"text-align: center;\">[latex]\\large{D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2}.[\/latex]<\/p>\n<ol id=\"fs-id1167793417260\" type=\"i\">\n<li>If\u00a0[latex]D>0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)>0[\/latex], then\u00a0[latex]f[\/latex] has a local minimum at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D>0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)<0[\/latex], then\u00a0[latex]f[\/latex] has a local maximum at\u00a0[latex](x_0,y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D<0[\/latex], then\u00a0[latex]f[\/latex] has a saddle point at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D=0[\/latex], then the test is inconclusive.<\/li>\n<\/ol>\n<p>See\u00a0Figure 2.<\/p>\n<\/div>\n<div id=\"attachment_1281\" style=\"width: 953px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1281\" class=\"size-full wp-image-1281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134549\/4-7-4.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. Figure a has two bulbous mounds pointing down, and the two extrema are listed as the local minima. Figure b has two bulbous mounds pointed up, and the two extrema are listed as the local maxima. Figure c is shaped like a saddle, and in the middle of the saddle, a point is marked as the saddle point.\" width=\"943\" height=\"345\" \/><\/p>\n<p id=\"caption-attachment-1281\" class=\"wp-caption-text\">Figure 2. The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).<\/p>\n<\/div>\n<p>To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.<\/p>\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\">Problem-solving strategy: using the second derivative test for functions of two variables<\/h3>\n<hr \/>\n<p>Let\u00a0[latex]z=f(x,y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. To apply the second derivative test to find local extrema, use the following steps:<\/p>\n<ol>\n<li>Determine the critical points [latex](x_0, y_0)[\/latex]\u00a0of the function [latex]f[\/latex] where [latex]f_x(x_0, y_0)=f_y(x_0, y_0)[\/latex]. Discard any points where at least one of the partial derivatives does not exist.<\/li>\n<li>Calculate the discriminant [latex]D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2[\/latex]\u00a0for each critical point of [latex]f[\/latex].<\/li>\n<li>Apply\u00a0Second Derivative Test\u00a0to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: using the second derivative test<\/h3>\n<p id=\"fs-id1167793244016\">Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:<\/p>\n<ol id=\"fs-id1167793374827\" type=\"a\">\n<li>[latex]f(x, y)=4x^{2}+9y^{2}+8x-36y+24[\/latex]<\/li>\n<li>[latex]g(x, y)=\\frac{1}{3}x^{3}+y^{2}+2xy-6x-3y+4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q914523760\">Show Solution<\/span><\/p>\n<div id=\"q914523760\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Step 1 of the problem-solving strategy involves finding the critical points of [latex]f[\/latex]. To do this, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set each of them equal to zero:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    f_x(x,y)&=8x+8 \\\\    f_y(x,y)&=18y-36.    \\end{align}[\/latex]<\/p>\n<p>Setting them equal to zero yields the system of equations<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    8x+8&=0 \\\\    18y-36&=0.    \\end{align}[\/latex]<\/p>\n<p>The solution to this system is [latex]x=-1[\/latex] and [latex]y=2[\/latex]. Therefore [latex](-1, 2)[\/latex] is a critical point of [latex]f[\/latex].<span data-type=\"newline\"><br \/>\n<\/span>Step 2 of the problem-solving strategy involves calculating [latex]D[\/latex]. To do this, we first calculate the second partial derivatives of [latex]f[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    f_{xx}(x,y)&=8 \\\\    f_{xy}(x,y)&=0 \\\\    f_{yy}(x,y)&=18.    \\end{align}[\/latex]<\/p>\n<p>Therefore, [latex]D=f_{xx}(-1,2)f_{yy}(-1,2)-(f_{xy}(-1,2))^2=(8)(18)-(0)^2=144[\/latex].<\/p>\n<p>Step 3 states to check the\u00a0Second Derivative Test for Functions of Two Variables.. Since [latex]D>0[\/latex] and [latex]f_{xx}(-1,2)>0[\/latex], this corresponds to case 1. Therefore, [latex]f[\/latex] has a local minimum at [latex](-1, 2)[\/latex] as shown in the following figure.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"attachment_1283\" style=\"width: 823px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1283\" class=\"wp-image-1283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135113\/4-7-5.jpeg\" alt=\"The function f(x, y) = 4x2 + 9y2 + 8x \u2013 36y + 24 is shown with local minimum at (\u20131, 2, \u201316). The shape is a plane curving up on both ends parallel to the y axis.\" width=\"813\" height=\"405\" \/><\/p>\n<p id=\"caption-attachment-1283\" class=\"wp-caption-text\">Figure 3. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum at [latex]\\small{(-1,2,-16)}[\/latex].<\/p>\n<\/div>\n<p>b. For step 1, we first calculate [latex]g_x(x,y)[\/latex] and [latex]g_y(x,y)[\/latex], then set each of them equal to zero:[latex]\\hspace{8cm}\\begin{align}g_x(x,y)&=x^2+2y-6 \\\\g_y(x,y)&=2y+2x-3.\\end{align}[\/latex]<br \/>\nSetting them equal to zero yields the system of equations<br \/>\n[latex]\\hspace{8cm}\\begin{align}x^2+2y-6&=0 \\\\2y+2x-3&=0.\\end{align}[\/latex]<br \/>\nTo solve this system, first solve the second equation for [latex]y[\/latex]. This gives [latex]y=\\frac{3-2x}{2}[\/latex]. Substituting this into the first equation gives[latex]\\hspace{7.5cm}\\begin{align}x^2+3-2x-6&=0 \\\\x^2-2x-3&=0 \\\\(x-3)(x+1)&=0.\\end{align}[\/latex]<br \/>\nTherefore, [latex]x=-1[\/latex] or [latex]x=3[\/latex]. Substituting these values into the equation\u00a0[latex]y=\\frac{3-2x}{2}[\/latex] yields the critical points [latex](-1,\\frac52)[\/latex] and [latex](3,-\\frac32)[\/latex].<\/p>\n<p id=\"fs-id1167794138811\">Step 2 involves calculating the second partial derivatives of [latex]g[\/latex]:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    g_{xx}(x,y)&=2x \\\\    g_{xy}(x,y)&=2 \\\\    g_{yy}(x,y)&=2.    \\end{align}[\/latex]<\/p>\n<p>Then, we find a general formula for\u00a0[latex]D[\/latex]:<\/p>\n<p>[latex]\\hspace{7cm}\\begin{align}    D&=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)-(g_{xy}(x_0,y_0))^2 \\\\    &=(2x_0)(2)-2^2 \\\\    &=4x_0-4.    \\end{align}[\/latex]<\/p>\n<p>Next, we substitute each critical point into this formula:<\/p>\n<p>[latex]\\hspace{7cm}\\begin{align}    D(-1,\\frac52)&=(2(-1))(2)-(2)^2=-4-4=-8 \\\\    D(3,-\\frac32)&=(2(3))(2)-(2)^2=12-4=8.    \\end{align}[\/latex]<\/p>\n<p>In step 3, we note that, applying the Second Derivative Test for Functions of Two Variables to point [latex](-1,\\frac{5}{2})[\/latex] leads to case 3, which means that [latex](-1,\\frac{5}{2})[\/latex]\u00a0is a saddle point. Applying the theorem to point [latex](3,-\\frac{3}{2})[\/latex]\u00a0leads to case 1, which means that [latex](3,-\\frac{3}{2})[\/latex] corresponds to a local minimum as shown in the following figure.<\/p>\n<div id=\"attachment_1284\" style=\"width: 839px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1284\" class=\"size-full wp-image-1284\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135248\/4-7-6.jpeg\" alt=\"The function f(x, y) = (1\/3)x3 + y2 + + 2xy \u2013 6x \u2013 3y + 4 is shown with local minimum at (3, \u20133\/2, \u201329\/4) and saddle point at (\u22121, 5\/2, 41\/12). The shape is a plane curving up on the corners near (4, 3) and (\u22122, \u22122).\" width=\"829\" height=\"488\" \/><\/p>\n<p id=\"caption-attachment-1284\" class=\"wp-caption-text\">Figure 4. The function [latex]\\small{g(x,y)}[\/latex] has a local minimum and a saddle point.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Use the second derivative to find the local extrema of the function<\/p>\n<p>[latex]f(x, y)=x^{3}+2xy-6x-4y^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q233578991\">Show Solution<\/span><\/p>\n<div id=\"q233578991\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](\\frac{4}{3},\\frac{1}{3})[\/latex] is a saddle point,\u00a0[latex](-\\frac{3}{2},-\\frac{3}{8})[\/latex] is a local maximum.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186168&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0IESE6fWxpQ&amp;video_target=tpm-plugin-0hltm37e-0IESE6fWxpQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.35_transcript.html\">transcript for \u201cCP 4.35\u201d here (opens in new window).<\/a><\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3936\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.35. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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