You find a point (a, b) where f_x(a,b) = 0 and f_y(a,b) = 0, and computing the Hessian gives D = f_xx·f_yy − (f_xy)² = −5. What can you conclude about this point?
AIt is a local minimum because D is negative
BIt is a local maximum because the discriminant is negative
CIt is a saddle point — the function has a local min in some directions and a local max in others through this point
DThe test is inconclusive; more information is needed to classify the point
D < 0 is the definitive sign of a saddle point. The function increases in some directions through (a, b) and decreases in others — like the geometry of a mountain pass. D > 0 is required for the point to be an extremum (with the sign of f_xx then distinguishing max from min). D = 0 is the inconclusive case. Options A and B confuse the sign rule: a negative D does not indicate a minimum or maximum of any kind.
Question 2 Multiple Choice
You want to find the global minimum of f(x, y) = x² − y² on the closed disk x² + y² ≤ 4. After setting f_x = f_y = 0, you find one interior critical point at the origin. What must you do next to guarantee you have found the global minimum?
AEvaluate f at the origin — since it is the only critical point, it must be the global minimum
BUse the second partials test to classify the origin, and if it is a minimum, it must be the global minimum
CAlso evaluate f along the boundary circle x² + y² = 4, compare all values, and take the smallest
DCheck for additional interior critical points using the Hessian's eigenvalues
For global extrema on a closed, bounded domain, you must compare the values at ALL candidates: interior critical points AND boundary points. The origin is actually a saddle point here (D = (2)(−2) − 0² = −4 < 0), so the global minimum lies on the boundary. On the boundary circle, f = x² − y² = x² − (4 − x²) = 2x² − 4, minimized at x = 0, giving f = −4. The boundary analysis is not optional — on a closed domain, global extrema can and often do occur on the boundary.
Question 3 True / False
Most critical point of a differentiable function f(x, y) — most point where f_x = 0 and f_y = 0 — is either a local maximum or a local minimum.
TTrue
FFalse
Answer: False
Saddle points are critical points that are neither local maxima nor local minima. At a saddle point, the function increases in some directions and decreases in others. A classic example is f(x, y) = x² − y², where the origin satisfies f_x = f_y = 0 but is a saddle: the x-axis slice shows a minimum, and the y-axis slice shows a maximum. The second partials test (using the discriminant D) is needed to classify critical points, and D < 0 identifies saddle points.
Question 4 True / False
The global maximum of a continuous function on a closed, bounded region might occur on the boundary rather than at a point where both partial derivatives are zero.
TTrue
FFalse
Answer: True
On a closed domain, boundary points are candidates for global extrema that are completely missed by setting partial derivatives to zero. Interior critical points are candidates only in the interior. A simple example: f(x, y) = x on the unit square [0,1]² has no interior critical points (f_x = 1 ≠ 0 everywhere), so the maximum of 1 is achieved on the boundary edge x = 1. The complete procedure always requires checking interior critical points AND performing a separate optimization on each boundary piece.
Question 5 Short Answer
What is the key new phenomenon that distinguishes critical point analysis for functions of two variables from the single-variable case, and how does the second partials test address it?
Think about your answer, then reveal below.
Model answer: In single-variable calculus, a critical point (f'(x) = 0) is classified simply as a max, min, or inflection point. In two variables, a third type of critical point appears: the saddle point, where the function has a local minimum in some cross-sectional directions and a local maximum in others. The second partials test addresses this by computing the discriminant D = f_xx·f_yy − (f_xy)², which measures the curvature behavior simultaneously in all directions. D > 0 means the surface curves the same way in every direction (bowl-shaped up or down = extremum); D < 0 means it curves in opposite directions in different cross-sections (saddle). This classification has no single-variable analogue.
The geometric intuition is crucial: in one variable, you're on a curve and a flat-slope point must be a turnaround. In two variables, you're on a surface, and a flat-slope point could be a peak, a valley, or a mountain pass where you're at the bottom of the pass in one direction and the top of the ridge in another. The Hessian matrix and its determinant D capture whether the surface behaves like a bowl (consistent curvature) or a saddle (opposing curvatures).