Questions: Second Partial Test and the Hessian

Question 1

At a critical point of f(x, y), you compute f_xx = −4, f_yy = −3, f_xy = 2. What type of critical point is this?

• A. Local minimum, because both f_xx and f_yy are negative
• B. Saddle point, because one second partial is larger in magnitude than the other
• C. Local maximum, because D = f_xx·f_yy − (f_xy)² = 8 > 0 and f_xx < 0
• D. The test is inconclusive because the mixed partial f_xy is nonzero

Answer: (C)

D = (−4)(−3) − (2)² = 12 − 4 = 8 > 0. Since D > 0, the surface curves the same way in all directions — it's either a bowl up or bowl down. The sign of f_xx determines which: f_xx = −4 < 0 means the surface is concave downward, so this is a local maximum. Option A is a common error: having both f_xx < 0 and f_yy < 0 is necessary but not sufficient for a maximum — you must also check that D > 0 to rule out a saddle.

Question 2

At a critical point, the Hessian determinant D = 0. What does the second partial derivative test tell you?

• A. The critical point is a saddle point, because zero determinant indicates indefiniteness
• B. The critical point is a local minimum, because zero is the boundary between positive and negative curvature
• C. The test is inconclusive — higher-order information is needed to classify the critical point
• D. The critical point is an inflection point, analogous to the single-variable case when f''(a) = 0

Answer: (C)

D = 0 is exactly the degenerate case where the second-order approximation is neither positive definite, negative definite, nor indefinite — it is semidefinite. The second partial test cannot distinguish between a local minimum, local maximum, and saddle in this case. Higher-order terms in the Taylor expansion are needed. For example, f(x,y) = x⁴ + y⁴ has D = 0 at the origin but a local minimum, while f(x,y) = x³ − y³ also has D = 0 there but a saddle.

Question 3

At a critical point where f_xx > 0 and f_yy > 0, the point should be a local minimum.

• T. True
• F. False

Answer: False

This is the most tempting misconception. f_xx > 0 and f_yy > 0 means the surface curves upward in the x- and y-directions individually, but the mixed partial f_xy can introduce a 'twist' that tilts the surface into a saddle shape in an oblique direction. For example, f(x,y) = x² + y² − 4xy has f_xx = 2 > 0, f_yy = 2 > 0, but f_xy = −4, giving D = 4 − 16 = −12 < 0 — a saddle point. The determinant D captures whether the curvature is consistent in ALL directions, not just along the axes.

Question 4

If the Hessian determinant D = f_xx·f_yy − (f_xy)² is negative at a critical point, the surface has a saddle shape there — curving upward in one direction and downward in another.

• T. True
• F. False

Answer: True

D < 0 means the quadratic form ax² + 2bxy + cy² (the local approximation) is indefinite — it takes both positive and negative values for different directions (x, y). Geometrically, the surface curves upward along one direction through the critical point and downward along another, creating the characteristic saddle shape. The negative determinant measures the 'imbalance' between the curvatures: f_xx·f_yy < (f_xy)², meaning the cross-curvature overwhelms the product of the axial curvatures.

Question 5

Explain geometrically why D < 0 indicates a saddle point rather than an extremum. What role does the cross-term f_xy play in the test?

Model answer: D = f_xx·f_yy − (f_xy)² tests whether the local quadratic approximation curves in the same direction in every direction through the critical point. If D > 0, both 'principal curvatures' have the same sign, so the surface is bowl-shaped. If D < 0, the cross-curvature term (f_xy)² is large enough to flip the curvature in some oblique direction, creating a saddle. The f_xy term measures how the slope in the x-direction changes as you move in the y-direction — a 'twist' — and if this twist is strong enough relative to the axial curvatures, the surface tilts into a saddle even when both f_xx and f_yy are positive.

Formally, the quadratic form f_xx·u² + 2f_xy·uv + f_yy·v² (the second-order Taylor approximation in direction (u,v)) is positive definite iff f_xx > 0 and D > 0 — meaning it's positive in every direction. When D < 0, the form is indefinite: there exist directions where it's positive (the surface rises) and directions where it's negative (the surface falls). This is exactly the definition of a saddle point.