Second partial derivative test - Wikipedia In cases 1 and 2, the requirement that fxx fyy − fxy2 is positive at (x, y) implies that fxx and fyy have the same sign there Therefore, the second condition, that fxx be greater (or less) than zero, could equivalently be that fyy or tr (H) = fxx + fyy be greater (or less) than zero at that point
MA 16020 { EXAM FORMULAS THE SECOND DERIVATIVE TEST l derivatives are continuous Let d = fxxfyy (fxy)2 and suppose (a; b) is a critical point of f If d(a; b) > 0 and fxx ; b) > 0, then f has a relative minimum at (a; b) If d(a; b) > 0 and fxx ; b) < 0, then f has a relative maximum at (a; b) d(a; b) < 0, then f has a saddle point a
Extrema of Functions of 2 Variables - math. etsu. edu where D = fxxfyy - ( fxy) 2 is called the discriminant of f (i e , expanding (3) will result in (2) ) If D ( p,q) > 0, then z " (0) has the same sign as fxx ( p,q) in all directions u = <m,n>, thus implying a maximum if fxx ( p,q) < 0 and a minimum if fxx ( p,q) > 0
MA213 The second derivative test and Lagrange Multipliers The second . . . 3 The complete test Let D = fxxfyy f2 xy nd D < 0 then it is a saddle point, meaning the directional de ivatives can be both positive and negative if we choose is a local minimum Moreover, in case D = 0, this is a co If D 0 and fxx or fyy is negative, then it is a local maximum Moreover, in case D = 0, this is a conditional maximum
Second Derivative Test - Expii The formula we use is D= fxxfyy+[fxy]2 First, we must take the second derivatives fxx, fyy, fxy, and fyx Note: fxy and fyx should always be the same In order to take the second derivative, we must first take the first derivatives: If f(x,y) = 3x4 +4y2, then fx(x,y) = 12x3 and fy(x,y) = 8y
math. ucdavis. edu For all other cases (for example, if D = 0) this test is INCONCLUSIVE This means other methods must be used to determine if the critical point determines a maximum value, minimum value, or saddle point
c14s9. DVI - East Tennessee State University from the signs of fxx and fxxfyy − f 2 xy at (a, b) Multiply both sides of the previous equation by fxx and rearrange the right-hand side to get xy > 0 at (a, b) then Q(0) < 0 for all suffi-ciently small nonzero values of h and k, and f has a local maximum value at (a, b)