2nd_derivative - MIT OpenCourseWare http:/ocw.mit.edu 18.02...

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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 .
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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; fI1(xo) < 0 + xo is a local maximum point. 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. fx(x, Y) = 0, 1. Find the critical points by solving the simultaneous equations 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.
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This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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2nd_derivative - MIT OpenCourseWare http:/ocw.mit.edu 18.02...

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