Second Derivative Test & Hessian - Multivariable Calculus

1. A function $f(x, y)$ has a critical point $P$ where $f_{xx}(P) = -5$ and $f_{xy}(P) = f_{yy}(P)$. Let $k = f_{xy}(P)$. For what values of $k$ does the Second Derivative Test classify the point $P$ as a local maximum?
   Answer: $-5 < k < 0$ For a critical point to be a local maximum, the Second Derivative Test requires two conditions: $D > 0$ and $f_{xx} < 0$. We are given $f_{xx}(P) = -5$, so the second condition ($f_{xx} < 0$) is already satisfied. We need to find the values of $k$ that make the discriminant $D$ positive. The discriminant is $D = f_{xx}f_{yy} - (f_{xy})^2$. At point $P$, we have $f_{xx} = -5$, $f_{yy} = k$, and $f_{xy} = k$. So, $D = (-5)(k) - (k)^2 = -5k - k^2$. We need to solve the inequality $D > 0$: $-5k - k^2 > 0$ Multiplying by -1 and reversing the inequality sign: $k^2 + 5k < 0$ $k(k+5) < 0$ The roots of $k(k+5)=0$ are $k=0$ and $k=-5$. The quadratic $k(k+5)$ is a parabola opening upwards, so it is negative between its roots. Therefore, the inequality holds for $-5 < k < 0$.
2. Consider the function $f(x, y) = e^{-x^2-y^2+2x-4y-5}$. Using the Second Derivative Test, classify the sole critical point of this function.
   Answer: Local maximum Since the exponential function $e^u$ is strictly increasing, the local extrema of $f(x, y)$ will occur at the same points as the local extrema of its exponent, $g(x, y) = -x^2-y^2+2x-4y-5$. We can analyze $g(x,y)$ to classify the critical point. First, find the critical point by setting the partial derivatives of $g$ to zero: $g_x = -2x + 2 = 0 \implies x = 1$ $g_y = -2y - 4 = 0 \implies y = -2$ The sole critical point is $(1, -2)$. Next, find the second partial derivatives of $g$: $g_{xx} = -2$ $g_{yy} = -2$ $g_{xy} = 0$ These are constant for all $(x,y)$. The discriminant is $D = g_{xx}g_{yy} - (g_{xy})^2 = (-2)(-2) - (0)^2 = 4$. Since $D = 4 > 0$ and $g_{xx} = -2 < 0$, the function $g(x,y)$ has a local maximum at $(1, -2)$. Therefore, $f(x, y)$ also has a local maximum at $(1, -2)$.
3. Consider the system where $f(x,y) = ax^2 + bxy + cy^2 + dx + ey + k$ has a critical point at $(1,-2)$ and the second derivative test gives $D = b^2 - 4ac = 16$. Which constraint on the coefficients ensures this critical point is a saddle point?
   Answer: The given information is insufficient because the problem notation confuses quadratic form discriminant with Hessian determinant The student has confused the discriminant of a quadratic form ($b^2 - 4ac$) with the Hessian determinant. For $f(x,y) = ax^2 + bxy + cy^2 + dx + ey + k$, the Hessian matrix is $H = \begin{pmatrix} 2a & b \\ b & 2c \end{pmatrix}$ with determinant $D = 4ac - b^2$. The condition $b^2 - 4ac = 16 > 0$ means $D = -16 < 0$, which would indicate a saddle point. However, the problem statement incorrectly identifies $b^2 - 4ac$ as the result of the 'second derivative test,' creating confusion. The correct interpretation requires recognizing this notational error.
4. A function $f(x,y)$ has the property that its Hessian matrix at a critical point $(a,b)$ satisfies $\det(H) = 0$ but $\text{tr}(H) \neq 0$. What can be concluded about the behavior of $f$ near $(a,b)$?
   Answer: The standard second derivative test is inconclusive, requiring analysis of higher-order derivatives When $\det(H) = 0$, the Hessian has at least one zero eigenvalue, making the standard second derivative test inconclusive. The condition $\text{tr}(H) \neq 0$ means the sum of eigenvalues is nonzero, so if one eigenvalue is zero, the other must be nonzero. However, this doesn't automatically determine the point's nature - it could still be a minimum, maximum, or saddle point depending on the behavior along the eigenvector corresponding to the zero eigenvalue, which requires examining higher-order terms. Choice A incorrectly assumes zero eigenvalue implies saddle point. Choice B is overly specific. Choice D confuses critical point classification with gradient computation.
5. For the function $p(x,y) = x^2 + xy + y^2 - 6x - 3y + 5$, after finding the critical point and computing the Hessian determinant, what additional step is necessary to complete the classification?
   Answer: Check the sign of $p_{xx}$ at the critical point to distinguish between minimum and maximum The critical point is found by solving $p_x = 2x + y - 6 = 0$ and $p_y = x + 2y - 3 = 0$, giving $(3,0)$. The Hessian has $p_{xx} = 2$, $p_{yy} = 2$, $p_{xy} = 1$, so $D = 4 - 1 = 3 > 0$. Since $D > 0$, we need to check $p_{xx}$ to determine the type: $p_{xx} = 2 > 0$ indicates a local minimum. Choice B is irrelevant for classification. Choice C misunderstands 'degenerate critical point'. Choice D refers to sufficient conditions for the test's validity, which are typically assumed.
6. A function $g(x,y)$ has a critical point at $P(a,b)$ where $g_{xx}(P) = -3$ and $g_{yy}(P) = 5$. What is the nature of this critical point?
   Answer: Saddle point The discriminant for the Second Derivative Test is $D = g_{xx}g_{yy} - (g_{xy})^2$. At the point $P$, we have $g_{xx}(P) = -3$ and $g_{yy}(P) = 5$. Let's compute the product $g_{xx}g_{yy} = (-3)(5) = -15$. So, the discriminant is $D = -15 - (g_{xy})^2$. Since $g_{xy}$ is a real-valued function, $(g_{xy})^2$ must be greater than or equal to zero. Therefore, $D = -15 - (g_{xy})^2 \le -15$. Because $D$ is always negative, regardless of the value of $g_{xy}(P)$, the critical point $P$ must be a saddle point. The information given is sufficient.
7. Let $H$ be the Hessian matrix of a function $f(x,y)$ at a critical point $P$. If $\det(H) > 0$ and $\mathrm{tr}(H) < 0$, where $\mathrm{tr}(H)$ is the trace of $H$, what is the nature of the point $P$?
   Answer: Local maximum The Hessian matrix is $H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}$. The determinant is $\det(H) = f_{xx}f_{yy} - f_{xy}^2 = D$. The trace is $\mathrm{tr}(H) = f_{xx} + f_{yy}$. We are given $D > 0$. This implies that the point is either a local minimum or a local maximum. It also implies that $f_{xx}f_{yy} > f_{xy}^2 \ge 0$, so $f_{xx}$ and $f_{yy}$ must have the same sign. We are also given $\mathrm{tr}(H) = f_{xx} + f_{yy} < 0$. Since $f_{xx}$ and $f_{yy}$ have the same sign and their sum is negative, they must both be negative. In the Second Derivative Test, the conditions $D > 0$ and $f_{xx} < 0$ indicate a local maximum. Therefore, $P$ is a local maximum.
8. Let $P$ be a critical point of a function $f(x,y)$ with continuous second partial derivatives. At point $P$, it is known that $f_{xx} = 2$, $f_{yy} = 8$, and $f_{xy} = -4$. Which of the following can be concluded about the point $P$ using the Second Derivative Test?
   Answer: The test is inconclusive. To classify the critical point $P$, we must compute the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$. Given the values at $P$: $f_{xx} = 2$, $f_{yy} = 8$, and $f_{xy} = -4$. $D = (2)(8) - (-4)^2 = 16 - 16 = 0$. When the discriminant $D$ is zero, the Second Derivative Test is inconclusive. It is not possible to determine from this test alone whether the point is a minimum, maximum, or saddle point. Distractor A is incorrect because the conditions $f_{xx}>0$ and $f_{yy}>0$ are not sufficient; the test also requires $D>0$. Distractor B incorrectly uses the trace of the Hessian. Distractor C incorrectly links the sign of $f_{xy}$ to the classification.
9. For $g(x,y) = e^{-(x^2+y^2)}(x^2 + 2xy + y^2 - 1)$, the critical point analysis reveals that the origin requires special attention. What makes the classification at $(0,0)$ particularly challenging?
   Answer: The Hessian determinant equals zero due to the quadratic form $(x+y)^2$, making the standard test inconclusive At (0,0): $g(0,0) = e^0(-1) = -1$ and $\nabla g(0,0) = (0,0)$, confirming it's critical. The quadratic term $x^2 + 2xy + y^2 = (x+y)^2$ creates a special structure. Computing the Hessian at (0,0): the second partial derivatives involve terms that make the determinant zero, rendering the standard second derivative test inconclusive. The behavior along the line $x+y=0$ versus other directions requires higher-order analysis. Choice A is incorrect - exponential functions are infinitely differentiable. Choice C is wrong - the function is smooth everywhere.
10. The Hessian matrix of a function $f(x, y)$ at a critical point $(a, b)$ has eigenvalues $\lambda_1 = -3$ and $\lambda_2 = 4$. What is the nature of the critical point $(a, b)$?
    Answer: A saddle point, because the eigenvalues have opposite signs. The determinant of the Hessian matrix, $D$, is the product of its eigenvalues. At the critical point $(a, b)$, $D = \lambda_1 \lambda_2 = (-3)(4) = -12$. According to the Second Derivative Test, if $D < 0$, the critical point is a saddle point. The condition that eigenvalues have opposite signs is equivalent to their product being negative, which means $D<0$. Therefore, $(a, b)$ is a saddle point. For a local maximum, both eigenvalues must be negative. For a local minimum, both must be positive. An inconclusive test corresponds to $D=0$, meaning at least one eigenvalue is zero.
11. How many local minima does the function $f(x, y) = x^4 + y^4 - 4xy + 1$ have?
    Answer: 2 First, find the critical points by setting the gradient of $f$ to zero. $f_x = 4x^3 - 4y = 0 \implies y = x^3$ $f_y = 4y^3 - 4x = 0 \implies x = y^3$ Substitute the first equation into the second: $x = (x^3)^3 = x^9$. This gives $x^9 - x = x(x^8 - 1) = 0$. The real solutions are $x=0, x=1, x=-1$. The corresponding critical points are $(0,0)$, $(1,1)$, and $(-1,-1)$. Next, apply the Second Derivative Test. The second partial derivatives are: $f_{xx} = 12x^2$, $f_{yy} = 12y^2$, $f_{xy} = -4$. The discriminant is $D(x,y) = (12x^2)(12y^2) - (-4)^2 = 144x^2y^2 - 16$. 1. At $(0,0)$: $D(0,0) = -16 < 0$. This is a saddle point. 2. At $(1,1)$: $D(1,1) = 144(1)^2(1)^2 - 16 = 128 > 0$. And $f_{xx}(1,1) = 12(1)^2 = 12 > 0$. This is a local minimum. 3. At $(-1,-1)$: $D(-1,-1) = 144(-1)^2(-1)^2 - 16 = 128 > 0$. And $f_{xx}(-1,-1) = 12(-1)^2 = 12 > 0$. This is also a local minimum. Thus, the function has two local minima.
12. The function $h(x,y) = x^4 + y^4 - 4x^2y^2 + x^2 + y^2$ has a critical point at the origin. To determine its nature, which approach gives the most definitive conclusion?
    Answer: The function can be rewritten as $(x^2-y^2)^2 + x^2 + y^2$, immediately showing it's a local minimum Computing the Hessian at (0,0): $h_{xx}(0,0) = 2$, $h_{yy}(0,0) = 2$, $h_{xy}(0,0) = 0$. The discriminant is $D = 4 > 0$ and $h_{xx} > 0$, suggesting a local minimum by the standard test. However, the key insight is recognizing the algebraic form: $h(x,y) = (x^2-y^2)^2 + x^2 + y^2$. Since both terms are non-negative and equal zero only at (0,0), this definitively proves (0,0) is a global minimum. Choice A is technically correct but misses the simpler approach. Choice B is incorrect since the Hessian is not degenerate ($D \neq 0$).
13. The profit $P$ from producing $x$ units of product A and $y$ units of product B is given by a function $P(x,y)$. At the production level $(x_0, y_0)$ that maximizes profit, the Hessian matrix of $P$ is evaluated. Which of the following properties must the Hessian matrix have at $(x_0, y_0)$, assuming the Second Derivative Test is conclusive?
    Answer: Its determinant is positive and its diagonal entries are negative. For a function of two variables to have a local maximum at a critical point, the Second Derivative Test requires two conditions to be met: 1. The discriminant $D = P_{xx}P_{yy} - (P_{xy})^2$ must be positive. The discriminant is the determinant of the Hessian matrix, so the determinant must be positive. 2. The second partial derivative $P_{xx}$ must be negative. If $D > 0$, then $P_{xx}P_{yy} > (P_{xy})^2 \ge 0$, which means $P_{xx}$ and $P_{yy}$ must have the same sign. Since $P_{xx}$ must be negative for a maximum, $P_{yy}$ must also be negative. Thus, the diagonal entries of the Hessian ($P_{xx}$ and $P_{yy}$) must be negative. Therefore, to maximize profit, the Hessian must have a positive determinant and negative diagonal entries.
14. A function $f(x,y)$ has a critical point at the origin $(0,0)$. At this point, $f_{xx}(0,0) = 2$ and $f_{yy}(0,0)=8$. The Second Derivative Test is inconclusive for this point. Which of the following could be the value of $f_{xy}(0,0)$?
    Answer: $4$ The Second Derivative Test is inconclusive when the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ is equal to zero. We are given the values of the second partial derivatives at the origin: $f_{xx}(0,0) = 2$ and $f_{yy}(0,0) = 8$. Let $c = f_{xy}(0,0)$. We set the discriminant to zero: $D = (2)(8) - c^2 = 0$. This simplifies to $16 - c^2 = 0$, which means $c^2 = 16$. The possible values for $c$ are $c=4$ and $c=-4$. Of the answer choices provided, only 4 is a possible value for $f_{xy}(0,0)$.
15. A function is defined by $f(x, y) = x^3 - 3x + y^3 + ky^2$, where $k$ is a real constant. This function has a critical point at $(1, 0)$. For which values of $k$ does the Second Derivative Test guarantee that $f$ has a local minimum at $(1, 0)$?
    Answer: $k > 0$ First, calculate the second partial derivatives of $f(x, y)$. $f_x = 3x^2 - 3 \implies f_{xx} = 6x$ $f_y = 3y^2 + 2ky \implies f_{yy} = 6y + 2k$ $f_{xy} = 0$ Next, evaluate these at the critical point $(1, 0)$: $f_{xx}(1, 0) = 6(1) = 6$ $f_{yy}(1, 0) = 6(0) + 2k = 2k$ $f_{xy}(1, 0) = 0$ The discriminant is $D(x, y) = f_{xx}f_{yy} - (f_{xy})^2$. At $(1, 0)$, we have: $D(1, 0) = (6)(2k) - (0)^2 = 12k$. For a local minimum, the Second Derivative Test requires $D > 0$ and $f_{xx} > 0$. We have $f_{xx}(1, 0) = 6 > 0$, so this condition is met. We also need $D > 0$, which means $12k > 0$, or $k > 0$.
16. The function $f(x, y) = y^2 - x^4$ has a critical point at $(0, 0)$. At this point, the Second Derivative Test is inconclusive because the discriminant $D=0$. By analyzing the function's behavior near $(0, 0)$, what is the nature of this critical point?
    Answer: A saddle point. The Second Derivative Test is inconclusive since $f_{xx}=0, f_{yy}=2, f_{xy}=0$ at $(0,0)$, so $D=0$. We must analyze the function's behavior near the critical point $(0,0)$. The value of the function at the critical point is $f(0,0) = 0$. Consider approaching the origin along different paths: 1. Along the $y$-axis (where $x=0$): $f(0, y) = y^2$. For any $y \neq 0$, $f(0, y) > 0$. This suggests a possible local minimum. 2. Along the $x$-axis (where $y=0$): $f(x, 0) = -x^4$. For any $x \neq 0$, $f(x, 0) < 0$. This suggests a possible local maximum. Since the function takes on values greater than $f(0,0)$ and values less than $f(0,0)$ in any arbitrarily small neighborhood of the origin, the point $(0,0)$ is a saddle point. The distractors represent incomplete analyses.

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