Questions: Second Partial Test for Local Extrema (Hessian)

Question 1

At a critical point of f(x,y), you find f_xx = 3, f_yy = 4, and f_xy = 5. What type of point is this?

• A. A local minimum, because f_xx > 0 and f_yy > 0
• B. A local maximum, because both second partials are positive
• C. A saddle point, because D = f_xx·f_yy − (f_xy)² = 12 − 25 = −13 < 0
• D. The test is inconclusive because D = 0

Answer: (C)

D = f_xx·f_yy − (f_xy)² = (3)(4) − (5)² = 12 − 25 = −13 < 0. When D < 0, the critical point is a saddle point regardless of the signs of the individual second partials. The large cross-term f_xy = 5 introduces enough twisting to create a saddle even though f_xx and f_yy are both positive. Option A is the classic error: seeing f_xx > 0 and f_yy > 0 and concluding local minimum without computing D.

Question 2

What does D = f_xx·f_yy − (f_xy)² geometrically measure at a critical point?

• A. The rate of change of the gradient vector
• B. The average curvature of the surface at the critical point
• C. Whether the Hessian's eigenvalues have the same sign (D > 0) or opposite signs (D < 0), determining if the surface curves the same way in all directions or curves up in some and down in others
• D. The distance between the critical point and the nearest saddle point

Answer: (C)

D is the determinant of the Hessian. Its sign reflects whether the Hessian's eigenvalues agree in sign. When D > 0, both eigenvalues are positive (local min) or both negative (local max) — the surface curves the same way in every direction. When D < 0, eigenvalues have opposite signs — the surface curves up in some directions and down in others, producing a saddle. The cross-term f_xy encodes twisting; when large relative to f_xx and f_yy, it forces eigenvalues to have opposite signs.

Question 3

If D = f_xx·f_yy − (f_xy)² = 0 at a critical point, the second partials test is inconclusive — the point could be a local min, local max, or saddle.

• T. True
• F. False

Answer: True

When D = 0, the Hessian is singular and the test provides no information. The critical point could be any type. Other methods — evaluating f near the point, higher-order analysis, or geometric inspection — are required to classify it.

Question 4

If f_xx > 0 and f_yy > 0 at a critical point of f(x,y), then the point should be a local minimum.

• T. True
• F. False

Answer: False

Not necessarily — you must also confirm that D = f_xx·f_yy − (f_xy)² > 0. Even when both axial second partials are positive, a sufficiently large cross-derivative f_xy can make D negative, producing a saddle point. The function curves upward along the coordinate axes but curves downward in a diagonal direction, creating a saddle.

Question 5

Why can a large f_xy value turn what appears to be a local minimum (f_xx > 0, f_yy > 0) into a saddle point?

Model answer: f_xx and f_yy measure curvature only along the x and y axes. f_xy measures the 'twisting' of the surface — how the slope in the x-direction changes as you move in the y-direction. When f_xy is large relative to f_xx and f_yy, the surface curves sharply in a diagonal direction not captured by the axial second partials. D = f_xx·f_yy − (f_xy)² going negative means the twisting overwhelms the axial curvature, creating a direction along which the function decreases despite the positive axial second partials.

This is the multivariable lesson that you can't just check curvature along each axis independently — the mixed partial captures cross-direction interactions that can reverse the sign of curvature in off-axis directions.