3. Suppose that z = f(x,y) is a function for which f(3,-1) = 16, А. 1,(3,-1) = 0, f ,(3,-1) = -7, and f(3,-1) = f ,(3,-1) = 1,(3,-1) = 0, f(3,-1) = -3, Does this data say that f(3,-1)% 16 is a maximum, minimum, or neit Briefly explain your reasoning. Suppose I had changed the mixed partial to f(3,-1) = f (3,-1) = 5. В. What would your answer be in this case?
3. Suppose that z = f(x,y) is a function for which f(3,-1) = 16, А. 1,(3,-1) = 0, f ,(3,-1) = -7, and f(3,-1) = f ,(3,-1) = 1,(3,-1) = 0, f(3,-1) = -3, Does this data say that f(3,-1)% 16 is a maximum, minimum, or neit Briefly explain your reasoning. Suppose I had changed the mixed partial to f(3,-1) = f (3,-1) = 5. В. What would your answer be in this case?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Make all obvious simplifications.
![### Problem Statement
**Question 8**
**A.** Suppose that \( z = f(x, y) \) is a function for which:
- \( f(3, -1) = 16 \),
- \( f_x(3, -1) = 0 \),
- \( f_y(3, -1) = 0 \),
- \( f_{xx}(3, -1) = -3 \),
- \( f_{yy}(3, -1) = -7 \), and
- \( f_{xy}(3, -1) = f_{yx}(3, -1) = 4 \).
Does this data say that \( f(3, -1) = 16 \) is a maximum, minimum, or neither? **Briefly** explain your reasoning.
**B.** Suppose I had changed the mixed partial to \( f_{xy}(3, -1) = f_{yx}(3, -1) = 5 \). What would your answer be in this case?
### Explanation
This question involves using the second derivative test for functions of two variables to determine the local extremum at a given point.
#### Second Derivative Test
1. Calculate the discriminant:
\[
D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2
\]
2. Evaluate at the point \( (3, -1) \):
- For Part A:
\[
D = (-3) \cdot (-7) - 4^2 = 21 - 16 = 5
\]
- Since \( D > 0 \) and \( f_{xx}(3, -1) < 0 \), \( f(3, -1) = 16 \) is a local maximum.
3. For Part B, substitute the changed mixed partials:
\[
D = (-3) \cdot (-7) - 5^2 = 21 - 25 = -4
\]
- Since \( D < 0 \), the point \( (3, -1) \) is a saddle point, meaning neither a local maximum nor a local minimum.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c560a1a-db76-4fd0-82f5-78bd443b8157%2Fbba94538-bf30-4b1d-9ae7-98bffd1a9ca2%2Fyjmsw1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Question 8**
**A.** Suppose that \( z = f(x, y) \) is a function for which:
- \( f(3, -1) = 16 \),
- \( f_x(3, -1) = 0 \),
- \( f_y(3, -1) = 0 \),
- \( f_{xx}(3, -1) = -3 \),
- \( f_{yy}(3, -1) = -7 \), and
- \( f_{xy}(3, -1) = f_{yx}(3, -1) = 4 \).
Does this data say that \( f(3, -1) = 16 \) is a maximum, minimum, or neither? **Briefly** explain your reasoning.
**B.** Suppose I had changed the mixed partial to \( f_{xy}(3, -1) = f_{yx}(3, -1) = 5 \). What would your answer be in this case?
### Explanation
This question involves using the second derivative test for functions of two variables to determine the local extremum at a given point.
#### Second Derivative Test
1. Calculate the discriminant:
\[
D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2
\]
2. Evaluate at the point \( (3, -1) \):
- For Part A:
\[
D = (-3) \cdot (-7) - 4^2 = 21 - 16 = 5
\]
- Since \( D > 0 \) and \( f_{xx}(3, -1) < 0 \), \( f(3, -1) = 16 \) is a local maximum.
3. For Part B, substitute the changed mixed partials:
\[
D = (-3) \cdot (-7) - 5^2 = 21 - 25 = -4
\]
- Since \( D < 0 \), the point \( (3, -1) \) is a saddle point, meaning neither a local maximum nor a local minimum.
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