Determine critical points of 2 variable functions without 2nd derivative test
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I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
asked Nov 22 at 6:56
p.chives
1
add a comment |
up vote
0
down vote
favorite
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
asked Nov 22 at 6:56
p.chives
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
asked Nov 22 at 6:56
p.chives
1
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
calculus
asked Nov 22 at 6:56
p.chives
1
asked Nov 22 at 6:56
p.chives
1
asked Nov 22 at 6:56
p.chives
1
asked Nov 22 at 6:56
p.chives
1
asked Nov 22 at 6:56
p.chives
1
1
add a comment |
add a comment |
2 Answers
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Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.5k2736
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Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
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2 Answers
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2 Answers
2
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up vote
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Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.5k2736
add a comment |
up vote
1
down vote
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.5k2736
add a comment |
up vote
1
down vote
up vote
1
down vote
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.5k2736
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.5k2736
answered Nov 22 at 7:24
farruhota
18.5k2736
answered Nov 22 at 7:24
farruhota
18.5k2736
answered Nov 22 at 7:24
farruhota
18.5k2736
18.5k2736
add a comment |
add a comment |
up vote
0
down vote
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
up vote
0
down vote
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
up vote
0
down vote
up vote
0
down vote
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
edited Nov 22 at 7:18
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
answered Nov 22 at 7:16
Mostafa Ayaz
13.5k3836
13.5k3836
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
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