Second derivative test for discriminant
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?
multivariable-calculus
edited Dec 2 '18 at 14:12
gimusi
1
asked Dec 2 '18 at 14:10
CCola
275
add a comment |
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?
multivariable-calculus
edited Dec 2 '18 at 14:12
gimusi
1
asked Dec 2 '18 at 14:10
CCola
275
add a comment |
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?
multivariable-calculus
edited Dec 2 '18 at 14:12
gimusi
1
asked Dec 2 '18 at 14:10
CCola
275
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?
multivariable-calculus
multivariable-calculus
edited Dec 2 '18 at 14:12
gimusi
1
asked Dec 2 '18 at 14:10
CCola
275
edited Dec 2 '18 at 14:12
gimusi
1
asked Dec 2 '18 at 14:10
CCola
275
edited Dec 2 '18 at 14:12
gimusi
1
edited Dec 2 '18 at 14:12
gimusi
1
edited Dec 2 '18 at 14:12
gimusi
1
1
asked Dec 2 '18 at 14:10
CCola
275
asked Dec 2 '18 at 14:10
CCola
275
asked Dec 2 '18 at 14:10
CCola
275
275
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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
answered Dec 2 '18 at 14:15
gimusi
1
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
answered Dec 2 '18 at 14:15
gimusi
1
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
add a comment |
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
answered Dec 2 '18 at 14:15
gimusi
1
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
add a comment |
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
answered Dec 2 '18 at 14:15
gimusi
1
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
answered Dec 2 '18 at 14:15
gimusi
1
edited Dec 2 '18 at 14:41
answered Dec 2 '18 at 14:15
gimusi
1
answered Dec 2 '18 at 14:15
gimusi
1
answered Dec 2 '18 at 14:15
gimusi
1
1
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
add a comment |
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
x should be equal to +/- root2?
– CCola
Dec 2 '18 at 14:38
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
@CCola Opsss yes of course...I fix the typo!
– gimusi
Dec 2 '18 at 14:41
add a comment |
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