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Second Partials Test Let F(X, Y) Be a Smooth Function and Let (X0,Y0)

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Second Partials Test Let F(X, Y) Be a Smooth Function and Let (X0,Y0) Multivariate Calculus Grinshpan Second Partials Test Let f(x; y) be a smooth function and let (x0; y0) be a point in its domain. It is possible to write f(x; y) = f(x0; y0) + fx(x0; y0)(x − x0) + fy(x0; y0)(y − y0) 1 2 1 2 + 2 fxx(x0; y0)(x−x0) +fxy(x0; y0)(x−x0)(y−y0)+ 2 fyy(x0; y0)(y−y0) + higher order terms: If (x0; y0) is a stationary point of f then df = 0 there and we have 1 2 1 2 f(x; y) = f(x0; y0) + 2 fxx(x0; y0)(x−x0) +fxy(x0; y0)(x−x0)(y−y0)+ 2 fyy(x0; y0)(y−y0) + higher order terms: Assume for simplicity that (x0; y0) = (0; 0) and f(0; 0) = 0. 0 0 0 0 (This can always be done by translating variables: f~(x ; y ) = f(x0 + x ; y0 + y ) − f(x0; y0).) Then letting 1 1 1 1 A = 2 fxx(0; 0);B = 2 fxy(0; 0) = 2 fyx(0; 0);C = 2 fyy(0; 0); we see that f(x; y) = Ax2 + 2Bxy + Cy2 + higher order terms: Thus the graph of f near (0; 0) resembles the quadric surface z = Ax2 + 2Bxy + Cy2: This quadric surface can either be an elliptic or a hyperbolic paraboloid. If it is an elliptic paraboloid opening up, (0; 0) is a point of local minimum. If it is an elliptic paraboloid opening down, (0; 0) is a point of local maximum. If it is a hyperbolic paraboloid, (0; 0) is a saddle point. It should certainly be possible to tell which case we are dealing with by looking at the coefficients A, B, and C, and this is the idea behind the Second Partials Test. Since, for y 6= 0, Ax2 + 2Bxy + Cy2 has the same sign as x 2 x A( y ) + 2B( y ) + C, it suffices to examine the one-variable quadratic At2 + 2Bt + C instead (this observation is repeatedly used below). Elliptic Paraboloid Opening Up. For any (x; y) different from (0; 0) we must have Ax2 + 2Bxy + Cy2 > 0: Setting x = 1; y = 0 we see that A > 0, setting x = 0; y = 1 we see that C > 0. Setting x = t; y = 1, we see that the parabola At2 + 2Bt + C does 2 not have real roots and so its discriminant 4(B2 − AC) is negative. Thus the conditions 2 fxx or fyy > 0; fxxfyy − fxy > 0 guarantee a local minimum. Elliptic Paraboloid Opening Down. For any (x; y) different from (0; 0) we must have Ax2 + 2Bxy + Cy2 < 0: Setting x = 1; y = 0 we see that A < 0, setting x = 0; y = 1 we see that C < 0. Setting x = t; y = 1, we see that the parabola At2 + 2Bt + C does not have real roots and so its discriminant 4(B2 − AC) is negative. Thus the conditions 2 fxx or fyy < 0; fxxfyy − fxy > 0 guarantee a local maximum. Hyperbolic Paraboloid. The expression Ax2 + 2Bxy + Cy2 can neither be positive nor negative for all (x; y) 6= (0; 0). This means that the parabola At2 + 2Bt + C takes both positive and negative values and so its discriminant 4(B2 − AC) is positive. Thus the condition 2 fxxfyy − fxy < 0 guarantees a saddle point. Not that this is the case if, for instance, A or C = 0 and B 6= 0, or if A and C are of opposite sign. We have been fairly informal in our analysis. In particular, we've left out the degenerate case B2 − AC = 0 where the higher-order terms of f (and so higher-order partials) become important. But the generic situation has been examined..
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