Questions: The Hessian Matrix and Second Derivative Test

Question 1

At a critical point of f(x, y), you compute det(H) = −3. What can you conclude?

• A. Local minimum, because the negative value indicates downward curvature
• B. Local maximum, because the Hessian determinant is negative
• C. Saddle point, because the Hessian has eigenvalues of opposite signs
• D. The test is inconclusive — you need higher-order information

Answer: (C)

det(H) < 0 means the Hessian has one positive and one negative eigenvalue — the function curves upward in some directions and downward in others. This is definitively a saddle point. A negative determinant rules out both local minima (which require all positive eigenvalues) and local maxima (all negative). The inconclusive case is det(H) = 0, where an eigenvalue is zero.

Question 2

At a critical point, f_xx = 4, f_yy = 2, and f_xy = 3. A student claims this must be a local minimum because f_xx > 0. Is the student correct?

• A. Yes — f_xx > 0 confirms upward curvature, guaranteeing a local minimum
• B. No — you must also verify f_yy > 0 to confirm a local minimum
• C. No — det(H) = (4)(2) − (3)² = −1 < 0, so this is a saddle point
• D. No — you need to compute both eigenvalues explicitly before concluding anything

Answer: (C)

det(H) = f_xx · f_yy − (f_xy)² = 8 − 9 = −1 < 0, so this is a saddle point. The student's error is the key misconception: f_xx > 0 only tells you the curvature in the x-direction is upward. A function can curve upward along x but downward in some other direction — that's a saddle. You need det(H) > 0 AND f_xx > 0 together to conclude local minimum.

Question 3

A saddle point occurs when the Hessian matrix has eigenvalues of opposite signs, which is equivalent to det(H) < 0.

• T. True
• F. False

Answer: True

At a saddle point, the function increases along some directions through the critical point and decreases along others. The Hessian captures this: one positive eigenvalue (upward curvature) and one negative eigenvalue (downward curvature). Since det(H) = λ₁ · λ₂, opposite-sign eigenvalues produce det(H) < 0. This is the test condition for a saddle point.

Question 4

If the Hessian determinant equals zero at a critical point, the point is expected to be a saddle point.

• T. True
• F. False

Answer: False

det(H) = 0 means at least one eigenvalue is zero — the test is inconclusive. The critical point could be a local minimum, a local maximum, or a saddle — you cannot determine which from the Hessian alone. Higher-order analysis or direct examination of the function's behavior is required. Confusing the inconclusive case (det = 0) with the saddle case (det < 0) is a common error.

Question 5

Why is checking f_xx > 0 alone insufficient to classify a critical point of a two-variable function as a local minimum?

Model answer: f_xx > 0 only tells you the function curves upward in the x-direction at that point. A function of two variables can curve upward along x and downward along some other direction — making the point a saddle, not a minimum. To guarantee a local minimum, the function must curve upward in every direction simultaneously, which requires the Hessian to be positive definite. This is captured by det(H) > 0 AND f_xx > 0 together.

The single-variable test works because there is only one direction. In two or more variables, upward curvature must hold along every direction through the critical point. The Hessian encodes curvature in all directions via its eigenvalues, and positive definiteness (all eigenvalues > 0) is the correct generalization of f''(c) > 0. Checking just f_xx is equivalent to checking curvature only along the x-axis and ignoring all others.