Studdy AI Logo
Math  /  Calculus

Question44. The discriminant fxxfyy−fxy2f_{x x} f_{y y}-f_{x y}{ }^{2} is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface z=f(x,y)z=f(x, y) looks like. Describe your reasoning in each case. a. f(x,y)=x2y2f(x, y)=x^{2} y^{2} b. f(x,y)=1−x2y2f(x, y)=1-x^{2} y^{2} c. f(x,y)=xy2f(x, y)=x y^{2} d. f(x,y)=x3y2f(x, y)=x^{3} y^{2} e. f(x,y)=x3y3f(x, y)=x^{3} y^{3} f. f(x,y)=x4y4f(x, y)=x^{4} y^{4}

Studdy Solution

STEP 1

1. The discriminant fxxfyy−fxy2 f_{xx} f_{yy} - f_{xy}^2 is zero at the origin for each function.
2. The Second Derivative Test fails at the origin.
3. We need to determine the nature of the critical point at the origin for each function by visualizing the surface z=f(x,y) z = f(x, y) .

STEP 2

1. Analyze each function individually to understand its behavior near the origin.
2. Determine the nature of the critical point at the origin for each function.

STEP 3

Consider f(x,y)=x2y2 f(x, y) = x^2 y^2 .
- At the origin, f(0,0)=0 f(0, 0) = 0 . - Away from the origin, f(x,y)≥0 f(x, y) \geq 0 for all x,y x, y . - The surface z=x2y2 z = x^2 y^2 is always non-negative and flattens out at the origin.
Conclusion: The origin is a minimum point because the function is non-negative and zero at the origin.

STEP 4

Consider f(x,y)=1−x2y2 f(x, y) = 1 - x^2 y^2 .
- At the origin, f(0,0)=1 f(0, 0) = 1 . - Away from the origin, f(x,y)≤1 f(x, y) \leq 1 . - The surface z=1−x2y2 z = 1 - x^2 y^2 is always less than or equal to 1 and peaks at the origin.
Conclusion: The origin is a maximum point because the function is at its highest value at the origin.

STEP 5

Consider f(x,y)=xy2 f(x, y) = x y^2 .
- At the origin, f(0,0)=0 f(0, 0) = 0 . - The function changes sign depending on x x (positive for x>0 x > 0 , negative for x<0 x < 0 ). - The surface z=xy2 z = x y^2 is a saddle surface.
Conclusion: The origin is neither a maximum nor a minimum point; it is a saddle point.

STEP 6

Consider f(x,y)=x3y2 f(x, y) = x^3 y^2 .
- At the origin, f(0,0)=0 f(0, 0) = 0 . - The function changes sign depending on x x (positive for x>0 x > 0 , negative for x<0 x < 0 ). - The surface z=x3y2 z = x^3 y^2 is a saddle surface.
Conclusion: The origin is neither a maximum nor a minimum point; it is a saddle point.

STEP 7

Consider f(x,y)=x3y3 f(x, y) = x^3 y^3 .
- At the origin, f(0,0)=0 f(0, 0) = 0 . - The function changes sign depending on the signs of x x and y y . - The surface z=x3y3 z = x^3 y^3 is a saddle surface.
Conclusion: The origin is neither a maximum nor a minimum point; it is a saddle point.

STEP 8

Consider f(x,y)=x4y4 f(x, y) = x^4 y^4 .
- At the origin, f(0,0)=0 f(0, 0) = 0 . - Away from the origin, f(x,y)≥0 f(x, y) \geq 0 for all x,y x, y . - The surface z=x4y4 z = x^4 y^4 is always non-negative and flattens out at the origin.
Conclusion: The origin is a minimum point because the function is non-negative and zero at the origin.
The nature of the critical points at the origin for each function is as follows: a. Minimum b. Maximum c. Neither (saddle point) d. Neither (saddle point) e. Neither (saddle point) f. Minimum

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord