Multivariable Calculus Quiz: Second Derivative Test And Hessian

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Second Derivative Test And HessianQuestion 1 of 16

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?

P

is a local minimum because

f_{xx} > 0

and

f_{yy} > 0

P

is a local maximum because the trace of the Hessian

f_{xx} + f_{yy}

is positive.

P

is a saddle point because

f_{xy}

is negative.

The test is inconclusive.

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Second Derivative Test And Hessian

Practice Second Derivative Test And Hessian in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Second Derivative Test And Hessian, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$P$ is a local minimum because $f_{xx} > 0$ and $f_{yy} > 0$ .
$P$ is a local maximum because the trace of the Hessian $f_{xx} + f_{yy}$ is positive.
$P$ is a saddle point because $f_{xy}$ is negative.
The test is inconclusive. (correct answer)

Explanation: 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.

Question 2

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 $f_{xy}(0,0)$ ?

$2$
$4$ (correct answer)
$8$
$16$

Explanation: 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)

Question 3

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?

$k > 0$ or $k < -5$
$k > 5$ or $k < 0$
$0 < k < 5$
$-5 < k < 0$ (correct answer)

Explanation: 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

Question 4

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?

A local minimum, because $f(x,y) \ge 0$ along the $y$ -axis.
A local maximum, because $f(x,y) \le 0$ along the $x$ -axis.
A saddle point. (correct answer)
The nature cannot be determined without a higher-order derivative test.

Explanation: The Second Derivative Test is inconclusive since

f_{xx}=0, f_{yy}=2, f_{xy}=0

(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:

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.
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.

Question 5

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 $(1, 0)$ ?

$k > 0$ (correct answer)
$k < 0$
$k = 0$
No values of $k$ exist.

Explanation: 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

Question 6

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 $P$ ?

Local maximum (correct answer)
Local minimum
Saddle point
The information is insufficient to classify the point.

Explanation: 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.

Question 7

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?

Its determinant is negative.
Its determinant is positive and its diagonal entries are positive.
Its determinant is positive and its diagonal entries are negative. (correct answer)
Its determinant is zero.

Explanation: 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:

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.
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.

Question 8

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 $(a,b)$ ?

The point is definitely a saddle point since one eigenvalue is zero and the other is nonzero
The point could be a local extremum if the nonzero eigenvalue and higher-order terms align appropriately
The standard second derivative test is inconclusive, requiring analysis of higher-order derivatives (correct answer)
The function has a degenerate critical point where the gradient computation fails

Explanation: 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.

Question 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?

The exponential factor creates discontinuities in the second partial derivatives at the origin
The Hessian determinant equals zero due to the quadratic form $(x+y)^2$ , making the standard test inconclusive (correct answer)
The function is not differentiable at the origin due to the product of exponential and polynomial terms
The critical point coincides with a zero of the polynomial factor, requiring specialized analysis techniques

Explanation: 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.

Question 10

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?

$b^2 > 4ac$ and $a > 0$ , since the discriminant condition dominates the classification
$b^2 - 4ac > 0$ with no additional constraints needed, as this always indicates a saddle point
$b^2 - 4ac = 16 > 0$ automatically makes it a saddle point regardless of individual coefficient signs
The given information is insufficient because the problem notation confuses quadratic form discriminant with Hessian determinant (correct answer)

Explanation: 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.

Question 11

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 $(a, b)$ ?

A local maximum, because at least one eigenvalue is negative.
A local minimum, because at least one eigenvalue is positive.
A saddle point, because the eigenvalues have opposite signs. (correct answer)
The test is inconclusive, because the sum of the eigenvalues is positive.

Explanation: 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.

Question 12

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?

Local maximum
Local minimum
Saddle point (correct answer)
The nature cannot be determined without knowing the value of $g_{xy}(P)$ .

Explanation: 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.

Question 13

How many local minima does the function $f(x, y) = x^4 + y^4 - 4xy + 1$ have?

0
1
2 (correct answer)
3

Explanation: 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

At $(0,0)$ : $D(0,0) = -16 < 0$ . This is a saddle point.
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.
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.

Question 14

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.

Local minimum
Local maximum (correct answer)
Saddle point
The test is inconclusive.

Explanation: 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)

Question 15

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?

Check the sign of $p_{xx}$ at the critical point to distinguish between minimum and maximum (correct answer)
Evaluate the function value at the critical point to determine if it's above or below the surface
Compute the gradient magnitude to ensure the critical point is not degenerate
Verify that higher-order partial derivatives exist and are continuous in a neighborhood

Explanation: 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.

Question 16

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?

The standard second derivative test is sufficient since the discriminant is non-zero
Higher-order derivatives must be examined because the Hessian is degenerate at the origin
The function can be rewritten as $(x^2-y^2)^2 + x^2 + y^2$ , immediately showing it's a local minimum (correct answer)
Directional derivatives along coordinate axes provide complete information about the critical point type

Explanation: 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