📝 Second partial derivative test (14 MCQs)

📖 Explanation: The discriminant is calculated as $D = f_{xx}f_{yy} - (f_{xy})^2$. Substituting the given values yields $D = (0)(-4) - 0^2 = 0$. When $D = 0$, the Second Partials Test provides no information about the nature of the critical point. The student incorrectly computed or interpreted $D$, confusing a zero discriminant with a negative one, which is a common error analysis scenario requiring careful verification of arithmetic and test conditions.

📖 Explanation: This question tests conceptual understanding that the discriminant $D$ depends on the interaction term $f_{xy}$. Even if both pure second derivatives are positive, a sufficiently large mixed partial can make $D < 0$, indicating a saddle point. This reflects real-world modeling where cross-effects dominate individual curvatures, requiring students to move beyond simplistic sign checks and understand the geometric meaning of the full Hessian determinant.

📖 Explanation: Elliptical contours typically suggest extrema, but non-uniform spacing along diagonals indicates asymmetric curvature. If spacing increases in one direction and decreases in another, the surface curves upward in one diagonal and downward in the orthogonal diagonal, characteristic of a saddle point. Thus, despite an initially elliptical appearance, the underlying geometry produces $D < 0$. This graph-based interpretation requires linking visual contour behavior to analytical test outcomes beyond textbook examples.

📖 Explanation: When $D > 0$, the critical point is either a local minimum or maximum, determined solely by the sign of $f_{xx}$ (or equivalently $f_{yy}$). Since $f_{xx} = -3 < 0$, the function is concave down in the x-direction and, due to $D > 0$, in all directions, confirming a local maximum. This direct recall question reinforces the precise condition linking discriminant sign and second derivative sign, correcting the common misconception that $D > 0$ alone suffices.

📖 Explanation: This application question requires computing $D = f_{xx}f_{yy} - (f_{xy})^2 = 16 - 25 = -9$. Despite both pure second derivatives being positive, the large mixed partial makes $D < 0$, indicating a saddle point. The colleague’s reasoning reflects a widespread misconception that individual convexity guarantees joint convexity. In optimization contexts, ignoring cross-terms can lead to erroneous conclusions about minima, emphasizing the necessity of the full discriminant in multi-variable decision-making.

📖 Explanation: At (0,0), $f_{xx} = f_{yy} = f_{xy} = 0$, so $D = 0$ and the test fails. However, rewriting $f = (x^2 - y^2)^2$ reveals it is always non-negative and zero when $x^2 = y^2$, including along lines $y = \pm x$. Since $f \geq 0$ everywhere and equals zero at the origin, it is a global minimum. This Olympiad-style problem combines test limitations with algebraic insight, showing that vanishing second derivatives require higher-order analysis.

📖 Explanation: Compute $D_A = (2)(2) - 1^2 = 3 > 0$ with $f_{xx} > 0$, so minimum. For B, $D_B = 4 - 9 = -5 < 0$, so saddle. This mixed-concepts question highlights how the same pure curvatures can yield opposite classifications based solely on interaction strength. It challenges the intuition that similar marginal behaviors imply similar joint behavior, reinforcing that multivariable extrema depend critically on cross-derivatives, not just univariate analogs.

📖 Explanation: Calculate $D = (-1)(-1) - (-1)^2 = 1 - 1 = 0$. Although both pure second derivatives suggest concavity, the discriminant is zero, rendering the Second Partials Test inconclusive. The point could be a maximum, minimum, or saddle—further analysis is needed. This error analysis question targets the mistake of assuming negative pure derivatives guarantee a maximum without verifying $D > 0$, a subtle but critical distinction in applied contexts like thermodynamics.

📖 Explanation: Hyperbolic contour patterns are the hallmark of saddle points, where the surface rises in one direction and falls in another. This directly corresponds to $D < 0$ in the Second Partials Test. Recognizing this visual signature allows quick classification without computation, linking graphical intuition to analytical criteria. This graph-based recall question ensures students connect geometric features to test outcomes, reinforcing spatial reasoning alongside symbolic manipulation.

📖 Explanation: When $D > 0$, the critical point is a local extremum, and its type is determined by the sign of either $f_{xx}$ or $f_{yy}$ (they share the same sign when $D > 0$). Since $f_{yy} = -2 < 0$, it must be a local maximum. The peer incorrectly associated negative curvature with a minimum. This error analysis question clarifies a fundamental rule often reversed by beginners, ensuring correct interpretation of test outputs.

📖 Explanation: Here, $D = (0)(0) - (2)^2 = -4 < 0$. Even though both pure second derivatives vanish, the nonzero mixed partial ensures $D < 0$, definitively indicating a saddle point. This counters the misconception that vanishing pure derivatives imply inconclusive results; the cross-term alone can determine the nature. This application question emphasizes that the discriminant captures interaction effects independent of marginal curvatures, crucial in transformed optimization problems.

📖 Explanation: When $D > 0$ and $f_{xx} > 0$, the Hessian is positive definite, meaning both eigenvalues are positive. This aligns with the Second Partials Test indicating a local minimum. Comparing methods deepens conceptual understanding: the discriminant sign and trace sign together determine eigenvalue signs. This mixed-concepts question bridges computational and linear algebra perspectives, showing equivalence between determinant-based and spectral approaches, which is essential for advanced multivariable analysis.

📖 Explanation: Compute $D = (3)(3) - 3^2 = 0$. Equal positive pure derivatives do not ensure $D > 0$; the mixed partial can exactly balance them. The student assumes symmetry implies definiteness, but here the Hessian is singular. This error analysis question exposes overgeneralization from symmetric cases, stressing that even perfectly balanced derivatives can yield degenerate Hessians requiring higher-order tests.

📖 Explanation: Compute second partials: $f_{xx} = 6x$, $f_{yy} = -6x$, $f_{xy} = -6y$. At (0,0), all are zero, so $D = 0$. The test fails, consistent with the monkey saddle’s higher-order nature. Unlike quadratic saddles, this cubic surface has three ascending and three descending regions. This Olympiad-style question links test failure to geometric complexity, illustrating that vanishing second derivatives signal non-quadratic behavior requiring alternative analysis methods beyond the standard test.