MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms SD Second Derivative Test 1 The Second Derivative Test We begin by recalling the situation for twice differentiable functions f x of one variable To find their local or relative maxima and minima we 1 find the critical points i e the solutions of fl x 0 2 apply the second derivative test to each critical point xo f1I xo 0 xo is a local minimum point xo is a local maximum point fI1 xo 0 The idea behind it is at xo the slope fl xo 0 if fl1 xo 0 then fl x is strictly increasing for x near xo so that the slope is negative to the left of xo and positive to the right which shows that xo is a minimum point The reasoning for the maximum point is similar If fl1 xo 0 the test fails and one has to investigate further by taking more derivatives or getting more information about the graph Besides being a maximum or minimum such a point could also be a horizontal point of inflection The analogous test for maxima and minima of functions of two variables f x y is a little more complicated since there are several equations to satisfy several derivatives to be taken into account and another important geometric possibility for a critical point namely a saddle point This is a local minimax point around such a point the graph of f x y looks like the central part of a saddle or the region around the highest point of a mountain pass In the neighborhood of a saddle point the graph of the function lies both above and below its horizontal tangent plane at the point Your textbook has illustrations The second derivative test for maxima minima and saddle points has two steps 1 Find the critical points by solving the simultaneous equations fx x Y 0 fy x y 0 Since a critical point xo yo 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 2 To test such a point to see if it is a local maximum or minimum point we calculate the three second derivatives at the point we use subscript 0 to denote evaluation at xO yo so for example f o f xo yo and denote the values by A B and C we are assuming the derivatives exist and are continuous Second derivative test be as in 1 Then AC B 0 AC B 0 Let xo yo be a critical point o f f x y and A B and C 0 or C 0 A 0 or C 0 AC B 0 A xO yo is a minimum point 30 yo is a maximum point XO YO is a saddle point If AC B 2 0 the test fails and more investigation is needed Note that if AC B 2 0 then AC 0 so that A and C must have the same sign 18 02 NOTES 2 Example 1 Find the critical points of w 12x2 y3 12xy and determine their type Solution We calculate the partial derivatives easily To find the critical points we solve simultaneously the equations w 0 and w 0 we get Thus there are two critical points 0 O and 1 2 To determine their type we use the second derivative test we have AC B2 144y 144 so that at O O we have AC B 2 144 so it is a saddle point at 1 2 we have AC B 2 144 and A 0 so it is a a minimum point A plot of the level curves is given at the right which confirms the above Note that the behavior of the level curves near the origin can be determined by using the approximation w z 12x2 12xy this shows the level curves near 0 O look like those of x x y which are hyperbolas with asymptotes x x y 0 i e the y axis x 0 and the diagonal line x y O 2 Justification for the Second derivative Test The test involves the quantity AC B2 In general whenever we see the expressions B 2 4AC or B 2 AC or their negatives we should suspect that the quadratic formula is behind it all in one of its standard forms the second is less familiar but is often used to get rid of the excess two s This is what is happening here We want to know whether near a critical point Po the graph of our function w f x y always stays on one side of its horizontal tangent plane Po is then a maximum or minimum point or whether it lies partly above and partly below the tangent plane Po is then a saddle point As we will see this is determined by how the graph of a quadratic function f x lies with respect to the x axis Here is the basic lemma Lemma For the quadratic function Ax2 2Bx C 5 6 AC B 0 0 or C 0 AC B 0 A 0 or C 0 A AC B O j j AX 2Bx C 0 for all x AX 2Bx C 0 for all x Ax2 2Bx C 0 for some x Ax2 2 B x C 0 for some x SD SECOND DERIVATIVE TEST 3 Proof of the Lemma To prove 5 we note that the quadratic formula in the form 4 shows that the zeros of Ax2 2Bx C are imaginary i e it has no real zeros Therefore its graph must lie entirely on one side of the x axis which side can be determined from either A or C since A O lim A x 2 2 B x m C O 5 03 Ax2 2Bx C whenx O 0 or C 0 the reasoning is analogous and proves 6 If on the other hand AC B 2 0 formula 4 shows the quadratic function has two real If A roots so that its parabolic graph crosses the x axis twice and hence lies partly above and partly below it This proves 7 Proof of the Second derivative Test in a special case The simplest function is a linear function w wo ax by but it does not in general have maximum or minimum points and its second derivatives are all zero The simplest functions to have interesting critical points are the quadratic functions which we write in the form the 2 s will be explained momentarily 1 w wo ax by Ax2 2Bxy Cy2 8 2 Such a function has in general a unique critical point which we will assume is 0 O this gives the function a special form which we can determine by evaluating its partial derivatives at 0 O w A w B Wyy C w o a wy o b The neat look of the above explains the and 2B in …
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