Find the critical points of each of the functions below and classify them as a local maximum, minimum, or neither
$f ( x , y ) = ln ( 2 + sin ( x y ) )$. Consider only the critical point $(0,0)$.
I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.
$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 $
For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?
calculus multivariable-calculus vector-analysis
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
add a comment |
$f ( x , y ) = ln ( 2 + sin ( x y ) )$. Consider only the critical point $(0,0)$.
I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.
$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 $
For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?
calculus multivariable-calculus vector-analysis
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
2
If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12
add a comment |
$f ( x , y ) = ln ( 2 + sin ( x y ) )$. Consider only the critical point $(0,0)$.
I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.
$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 $
For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?
calculus multivariable-calculus vector-analysis
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
$f ( x , y ) = ln ( 2 + sin ( x y ) )$. Consider only the critical point $(0,0)$.
I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.
$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 $
For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?
calculus multivariable-calculus vector-analysis
calculus multivariable-calculus vector-analysis
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
asked Nov 12 '18 at 18:07
Mohammed Shahid
1457
1457
2
If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12
add a comment |
2
If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12
2
2
If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12
If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12
add a comment |
2 Answers
2
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oldest
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For the first one we have
$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=fracxy2+o(x^2+y^2)$$
and then it is a saddle point.
For the second one we have
$$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 ge left( x ^ 2 + y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 =r^2e^1-r^2=er^2-r^2+o(r^2)$$
that is positive as $rto 0$ and therefore it is a minimum.
answered Nov 12 '18 at 18:16
gimusi
1
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
|
show 4 more comments
Sometimes a graph is worth a thousand words:
As @gimusi points out, this is clearly a saddle point.
Here's your second function:
Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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For the first one we have
$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=fracxy2+o(x^2+y^2)$$
and then it is a saddle point.
For the second one we have
$$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 ge left( x ^ 2 + y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 =r^2e^1-r^2=er^2-r^2+o(r^2)$$
that is positive as $rto 0$ and therefore it is a minimum.
answered Nov 12 '18 at 18:16
gimusi
1
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
|
show 4 more comments
For the first one we have
$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=fracxy2+o(x^2+y^2)$$
and then it is a saddle point.
For the second one we have
$$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 ge left( x ^ 2 + y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 =r^2e^1-r^2=er^2-r^2+o(r^2)$$
that is positive as $rto 0$ and therefore it is a minimum.
answered Nov 12 '18 at 18:16
gimusi
1
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
|
show 4 more comments
For the first one we have
$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=fracxy2+o(x^2+y^2)$$
and then it is a saddle point.
For the second one we have
$$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 ge left( x ^ 2 + y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 =r^2e^1-r^2=er^2-r^2+o(r^2)$$
that is positive as $rto 0$ and therefore it is a minimum.
answered Nov 12 '18 at 18:16
gimusi
1
For the first one we have
$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=fracxy2+o(x^2+y^2)$$
and then it is a saddle point.
For the second one we have
$$f ( x , y ) = left( x ^ 2 + 3 y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 ge left( x ^ 2 + y ^ 2 right) e ^ 1 - x ^ 2 - y ^ 2 =r^2e^1-r^2=er^2-r^2+o(r^2)$$
that is positive as $rto 0$ and therefore it is a minimum.
answered Nov 12 '18 at 18:16
gimusi
1
edited Nov 12 '18 at 18:51
answered Nov 12 '18 at 18:16
gimusi
1
answered Nov 12 '18 at 18:16
gimusi
1
answered Nov 12 '18 at 18:16
gimusi
1
1
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
|
show 4 more comments
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part?
– Mohammed Shahid
Nov 12 '18 at 18:21
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
@MohammedShahid For the first function $f(0,0)=ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$.
– gimusi
Nov 12 '18 at 18:23
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $fracxy2+o(sqrtx^2+y^2)$ imply that there is a saddle point? Also, what does the $o(sqrtx^2+y^2)$ mean?
– Mohammed Shahid
Nov 12 '18 at 18:35
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
Are you aware about Taylor's expansion or at least first order expansion?
– gimusi
Nov 12 '18 at 18:36
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
I know about Taylor series that give polynomial approximation of a function. Is this similar?
– Mohammed Shahid
Nov 12 '18 at 18:41
|
show 4 more comments
Sometimes a graph is worth a thousand words:
As @gimusi points out, this is clearly a saddle point.
Here's your second function:
Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
add a comment |
Sometimes a graph is worth a thousand words:
As @gimusi points out, this is clearly a saddle point.
Here's your second function:
Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
add a comment |
Sometimes a graph is worth a thousand words:
As @gimusi points out, this is clearly a saddle point.
Here's your second function:
Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
Sometimes a graph is worth a thousand words:
As @gimusi points out, this is clearly a saddle point.
Here's your second function:
Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
edited Nov 12 '18 at 18:33
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
answered Nov 12 '18 at 18:20
David G. Stork
9,93521232
9,93521232
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
add a comment |
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
1
1
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
That's confirm also the second one is a minimum! Thanks
– gimusi
Nov 12 '18 at 18:24
add a comment |
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If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point.
– Doug M
Nov 12 '18 at 18:12