Find the critical points of each of the functions below and classify them as a local maximum, minimum, or...












3














$f ( x , y ) = ln ( 2 + sin ( x y ) )$. 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







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  • 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


















3














$f ( x , y ) = ln ( 2 + sin ( x y ) )$. 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







share|cite|improve this question


















  • 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
















3












3








3







$f ( x , y ) = ln ( 2 + sin ( x y ) )$. 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







share|cite|improve this question













$f ( x , y ) = ln ( 2 + sin ( x y ) )$. 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 12 '18 at 18:07









Mohammed ShahidMohammed 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
















  • 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












2 Answers
2






active

oldest

votes


















2














For the first one we have



$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=frac{xy}2+o(x^2+y^2)$$



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.






share|cite|improve this answer























  • 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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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



















3














Sometimes a graph is worth a thousand words:



enter image description here



As @gimusi points out, this is clearly a saddle point.



Here's your second function:



enter image description here



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.






share|cite|improve this answer



















  • 1




    That's confirm also the second one is a minimum! Thanks
    – gimusi
    Nov 12 '18 at 18:24











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














For the first one we have



$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=frac{xy}2+o(x^2+y^2)$$



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.






share|cite|improve this answer























  • 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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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
















2














For the first one we have



$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=frac{xy}2+o(x^2+y^2)$$



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.






share|cite|improve this answer























  • 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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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














2












2








2






For the first one we have



$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=frac{xy}2+o(x^2+y^2)$$



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.






share|cite|improve this answer














For the first one we have



$$f ( x , y ) -f(0,0)= ln ( 2 + sin ( x y ) )-ln 2=frac{xy}2+o(x^2+y^2)$$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 12 '18 at 18:51

























answered Nov 12 '18 at 18:16









gimusigimusi

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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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












  • @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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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 $frac{xy}{2}+o(sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(sqrt{x^2+y^2})$ 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











3














Sometimes a graph is worth a thousand words:



enter image description here



As @gimusi points out, this is clearly a saddle point.



Here's your second function:



enter image description here



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.






share|cite|improve this answer



















  • 1




    That's confirm also the second one is a minimum! Thanks
    – gimusi
    Nov 12 '18 at 18:24
















3














Sometimes a graph is worth a thousand words:



enter image description here



As @gimusi points out, this is clearly a saddle point.



Here's your second function:



enter image description here



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.






share|cite|improve this answer



















  • 1




    That's confirm also the second one is a minimum! Thanks
    – gimusi
    Nov 12 '18 at 18:24














3












3








3






Sometimes a graph is worth a thousand words:



enter image description here



As @gimusi points out, this is clearly a saddle point.



Here's your second function:



enter image description here



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.






share|cite|improve this answer














Sometimes a graph is worth a thousand words:



enter image description here



As @gimusi points out, this is clearly a saddle point.



Here's your second function:



enter image description here



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 12 '18 at 18:33

























answered Nov 12 '18 at 18:20









David G. StorkDavid G. Stork

10.1k21332




10.1k21332








  • 1




    That's confirm also the second one is a minimum! Thanks
    – gimusi
    Nov 12 '18 at 18:24














  • 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


















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