At a critical point of f(x,y), you compute f_xx = -3, f_yy = -2, f_xy = 1. What is the correct classification?
ASaddle point, because f_xx and f_yy have the same sign
BLocal maximum, because D = f_xx·f_yy − (f_xy)² = 5 > 0 and f_xx < 0
CLocal minimum, because both f_xx and f_yy are negative
DInconclusive, because D > 0 alone is insufficient without knowing f_xy's sign
D = (-3)(-2) − (1)² = 6 − 1 = 5 > 0, so curvature is consistent in all directions (the Hessian is definite). Since f_xx = -3 < 0, the surface curves downward in the x-direction, so it is a local maximum — not a minimum. Option C is the most tempting wrong answer: students see that both diagonal entries are negative and conclude minimum, forgetting that the correct procedure is to check the sign of f_xx after confirming D > 0.
Question 2 Multiple Choice
At a critical point of g(x,y), the discriminant D = f_xx·f_yy − (f_xy)² = 0. A student concludes: 'D = 0 means the Hessian is singular, so this must be a saddle point.' Is this correct?
AYes — a singular Hessian always corresponds to a saddle point in two variables
BNo — D = 0 is genuinely inconclusive; the critical point could be a local max, local min, or saddle
CNo — D = 0 means the test is inconclusive, but in practice it always indicates a flat region, not an extremum
DYes — D = 0 means one principal curvature is zero, ruling out strict extrema
D = 0 is genuinely inconclusive — all three outcomes (local max, local min, saddle point) are possible, and examples can be constructed for each. The function f(x,y) = x⁴ + y⁴ has a local minimum at (0,0) with D = 0; g(x,y) = x³ has a saddle at (0,0) with D = 0; h(x,y) = −x⁴ − y⁴ has a local max with D = 0. When D = 0, higher-order behavior must be examined. The student's reasoning — that a singular Hessian implies a saddle — is a common but incorrect inference.
Question 3 True / False
At a critical point where D > 0, the sign of f_yy can be used instead of f_xx to determine whether the point is a local max or min, and it will give the same classification.
TTrue
FFalse
Answer: True
When D > 0, the Hessian is positive definite (f_xx > 0 and f_yy > 0 → local min) or negative definite (f_xx < 0 and f_yy < 0 → local max). Because D > 0 forces f_xx and f_yy to have the same sign (if they differed in sign, D = f_xx·f_yy − (f_xy)² ≤ f_xx·f_yy < 0), checking f_yy gives the same answer as checking f_xx. The convention to use f_xx is just that — a convention, not a mathematical requirement.
Question 4 True / False
At a critical point where D < 0, the second partials test is inconclusive — more information is needed to determine whether the point is a max, min, or saddle.
TTrue
FFalse
Answer: False
D < 0 is conclusive: it guarantees a saddle point. A negative discriminant means the Hessian is indefinite — the surface curves upward in some directions and downward in others. You can always find two directions through the critical point where one cross-section is concave up and the other is concave down. This incompatibility between directions is precisely what defines a saddle point. The genuinely inconclusive case is D = 0, not D < 0.
Question 5 Short Answer
Explain geometrically why D < 0 at a critical point guarantees a saddle point rather than a local extremum.
Think about your answer, then reveal below.
Model answer: D < 0 means the Hessian matrix is indefinite — its eigenvalues have opposite signs. Geometrically, this means the surface has positive curvature (concave up) in some directions through the critical point and negative curvature (concave down) in others. A local minimum requires the surface to curve upward in every direction; a local maximum requires it to curve downward in every direction. When curvature changes sign depending on direction, neither condition is met, and the critical point must be a saddle — a point that is a local minimum along some cross-sections and a local maximum along others.
The discriminant D = f_xx·f_yy − (f_xy)² is the determinant of the Hessian, which is the product of its eigenvalues. A negative determinant means the eigenvalues have opposite signs — precisely the mathematical statement that curvature is not consistent in all directions. The geometric picture — slicing the surface at various angles through the critical point and observing that some slices curve up while others curve down — makes this intuitively clear.