fx = 3 x^2 + 4 y
level if x = +/-(4y/3)^.5
fxx = 6 x
fxy = 4
fy = 4 x - 4 y level when x = y
fyy = -4
fyx = 4
extreme only when x = y = 0
H matrix
6 x 4
4 -4
det = -6x - 4
if x = 0, det = -4
if det <0 then (0,0) is a "saddle point"
Using the Second Partials Test
Find the relative extrema of f(x,y)=-x^(3)+4xy-2y^(2)+1
1 answer