Second derivative test for discriminant

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Question: Find all critical points of $f(x,y)=x^3+y^4-6x-2y^2$

Apply 2nd Derivative test to each point and determine whether it is local maximum, local minimum or saddle point or that the test fails.

I have found six critical points in total and applying the discriminant equation for 2 variables [ discriminant= (fxx)(fyy) - (fxy)^2 , which is 24x(3y^2 -1) in this question ] , what should be my next step?

edited Dec 2 at 14:12

gimusi

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asked Dec 2 at 14:10

CCola

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up vote
1
down vote

favorite

Question: Find all critical points of $f(x,y)=x^3+y^4-6x-2y^2$

Apply 2nd Derivative test to each point and determine whether it is local maximum, local minimum or saddle point or that the test fails.

edited Dec 2 at 14:12

gimusi

90.6k74495

asked Dec 2 at 14:10

CCola

133

New contributor

add a comment |

up vote
1
down vote

favorite

Question: Find all critical points of $f(x,y)=x^3+y^4-6x-2y^2$

Apply 2nd Derivative test to each point and determine whether it is local maximum, local minimum or saddle point or that the test fails.

edited Dec 2 at 14:12

gimusi

90.6k74495

asked Dec 2 at 14:10

CCola

133

New contributor

Question: Find all critical points of $f(x,y)=x^3+y^4-6x-2y^2$

Apply 2nd Derivative test to each point and determine whether it is local maximum, local minimum or saddle point or that the test fails.

multivariable-calculus

edited Dec 2 at 14:12

gimusi

90.6k74495

asked Dec 2 at 14:10

CCola

133

New contributor

edited Dec 2 at 14:12

gimusi

90.6k74495

asked Dec 2 at 14:10

CCola

133

New contributor

edited Dec 2 at 14:12

gimusi

90.6k74495

edited Dec 2 at 14:12

gimusi

90.6k74495

edited Dec 2 at 14:12

gimusi

90.6k74495

asked Dec 2 at 14:10

CCola

133

New contributor

asked Dec 2 at 14:10

CCola

133

asked Dec 2 at 14:10

CCola

133

New contributor

CCola is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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HINT

We have that for $f(x,y)=x^3+y^4-6x-2y^2$

$f_x=3x^2-6=0 implies x=pm sqrt 2$

$f_y=4y^3-4y=0 implies y=0 ,lor,y=pm 1$

then evaluate

$f_{xx}=6x$

$f_{xy}=f_{yx}=0$

$f_{yy}=12y^2-4$

at each critical point and consider the Hessian determinant at each point.

Refer to

Second partial derivative test

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

accepted

HINT

We have that for $f(x,y)=x^3+y^4-6x-2y^2$

$f_x=3x^2-6=0 implies x=pm sqrt 2$

$f_y=4y^3-4y=0 implies y=0 ,lor,y=pm 1$

then evaluate

$f_{xx}=6x$

$f_{xy}=f_{yx}=0$

$f_{yy}=12y^2-4$

at each critical point and consider the Hessian determinant at each point.

Refer to

Second partial derivative test

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

add a comment |

up vote
0
down vote

accepted

HINT

We have that for $f(x,y)=x^3+y^4-6x-2y^2$

$f_x=3x^2-6=0 implies x=pm sqrt 2$

$f_y=4y^3-4y=0 implies y=0 ,lor,y=pm 1$

then evaluate

$f_{xx}=6x$

$f_{xy}=f_{yx}=0$

$f_{yy}=12y^2-4$

at each critical point and consider the Hessian determinant at each point.

Refer to

Second partial derivative test

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

add a comment |

up vote
0
down vote

accepted

HINT

We have that for $f(x,y)=x^3+y^4-6x-2y^2$

$f_x=3x^2-6=0 implies x=pm sqrt 2$

$f_y=4y^3-4y=0 implies y=0 ,lor,y=pm 1$

then evaluate

$f_{xx}=6x$

$f_{xy}=f_{yx}=0$

$f_{yy}=12y^2-4$

at each critical point and consider the Hessian determinant at each point.

Refer to

Second partial derivative test

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

HINT

We have that for $f(x,y)=x^3+y^4-6x-2y^2$

$f_x=3x^2-6=0 implies x=pm sqrt 2$

$f_y=4y^3-4y=0 implies y=0 ,lor,y=pm 1$

then evaluate

$f_{xx}=6x$

$f_{xy}=f_{yx}=0$

$f_{yy}=12y^2-4$

at each critical point and consider the Hessian determinant at each point.

Refer to

Second partial derivative test

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

edited Dec 2 at 14:41

answered Dec 2 at 14:15

gimusi

90.6k74495

answered Dec 2 at 14:15

gimusi

90.6k74495

answered Dec 2 at 14:15

gimusi

90.6k74495

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

add a comment |

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

x should be equal to +/- root2?
– CCola
Dec 2 at 14:38

@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 at 14:41

add a comment |

CCola is a new contributor. Be nice, and check out our Code of Conduct.

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