Questions: Critical Points, Extrema, and Saddle Points

Question 1

For f(x, y) = x² − y², you compute ∇f(0,0) = (0, 0). What type of critical point is the origin?

• A. A local minimum — the function curves upward near the origin
• B. A local maximum — the function curves downward near the origin
• C. A saddle point — the function increases in some directions and decreases in others
• D. Not a critical point — the gradient must be nonzero to classify it

Answer: (C)

Along the x-axis (y=0), f = x² increases away from the origin — like a valley. Along the y-axis (x=0), f = −y² decreases away from the origin — like a ridge. Since f is larger than f(0,0) in some directions and smaller in others, the origin is a saddle point, not an extremum. Option D is wrong: ∇f = 0 is exactly the condition defining a critical point. The origin is a critical point; the issue is that ∇f = 0 alone doesn't tell you what kind.

Question 2

Why does the saddle point phenomenon have no analogue in single-variable calculus?

• A. Single-variable functions can have saddle points, but they are called inflection points instead
• B. In one dimension, there is only one direction to move from a critical point, so the function either increases or decreases — there are no competing directions for a saddle to exist
• C. Single-variable calculus uses a different definition of critical point that excludes saddles
• D. Saddle points only occur when the Hessian has negative eigenvalues, which is impossible in 1D

Answer: (B)

A saddle point requires the function to increase in some directions and decrease in others. In one dimension, 'direction' means only left or right. If f'(c) = 0 at a critical point, the second derivative test tells you whether the function curves up (minimum), curves down (maximum), or has a degenerate case (inflection point at a horizontal tangent). But there are no competing spatial directions — the 'saddle' structure requires at least two independent directions. Option A conflates inflection points (where curvature changes sign in 1D) with saddle points (a multidimensional phenomenon).

Question 3

If ∇f(p) = 0 at a point p, then p is expected to be either a local maximum or a local minimum.

• T. True
• F. False

Answer: False

∇f = 0 is a necessary condition for a local extremum but not a sufficient one. Saddle points also satisfy ∇f = 0 yet are neither maxima nor minima. The canonical example is f(x,y) = x² − y² at the origin: the gradient vanishes, but the function increases along the x-axis and decreases along the y-axis. To determine which case applies, you need additional information about the curvature — provided by the Hessian matrix and the second-derivative test.

Question 4

Every local minimum of a differentiable function f: ℝ² → ℝ is a critical point.

• T. True
• F. False

Answer: True

This is the multivariable generalization of Fermat's theorem: if f has a local extremum at an interior point and is differentiable there, then ∇f = 0 at that point. The reasoning is that each partial derivative must be zero — if ∂f/∂x ≠ 0, then moving in the x-direction from the point would increase or decrease f, contradicting it being a local minimum. So ∇f = 0 is necessary (but not sufficient) for a local extremum.

Question 5

Why is finding all points where ∇f = 0 not sufficient to identify the minima of a function? What additional step is required?

Model answer: Solving ∇f = 0 gives a list of critical point candidates — but each candidate could be a local minimum, local maximum, or saddle point. To classify each one, you need the Hessian matrix (the matrix of second-order partial derivatives), which captures the curvature of the surface in every direction. The second-derivative test uses the Hessian's determinant and leading entry to distinguish these cases: positive definite Hessian → local minimum; negative definite → local maximum; indefinite → saddle point.

This distinction is practically crucial in optimization: an algorithm that stops at ∇f = 0 may have found a saddle rather than a minimum. In high-dimensional problems like neural network training, saddle points are extremely common and can stall gradient-based optimizers. The Hessian test is the theoretical tool; in practice, other methods (second-order optimization, checking the sign of the function change) are used to verify that a critical point is truly a minimum.