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Will someone please help me in the following?

I am given with the function $f(x,y)=(x+y)^3\sqrt{x^2+y^2}-1+\cos(x+y)$ and need to determine whether $(0,0)$ and $(1,-1)$ are extremum points or not.

As for $(0,0)$ after using Taylor expansion for $\cos(x+y)$ around $(0,0)$, we get that $f(x,y)=-\frac{(x+y)^2}{2}+O((x+y)^3) $ and hence we have that $(0,0)$ is a local maximum.

As for $(1,-1)$ I think that the same argument will work here as well, but I am not sure about it, since these are two different parts of the question.

Am I right? If not, where is my mistake?

Hope you'll be able to help.

Thanks in advance

homogenity
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  • For $(1,-1)$, you must first expand the function into a taylor series around $(1,-1)$. – 5xum May 14 '14 at 07:07
  • But why ? I am using Taylor expansion for $cos(x) $ around $x=0$ . It is the same if the point is $(1,-1) $ or if the point is $(0,0)$, right? Thanks ! – homogenity May 14 '14 at 07:09
  • Because if $x$ is close to $1$, then you cannot conclude that $x^3$ is very small. You can conclude, however, that $(x-1)^3$ is very small. Thus, your function must be something like $f(x,y)=f(1,-1) + ((x-1)^k + (y+1)^l)$. – 5xum May 14 '14 at 07:11
  • But $x$ is not close to 1... In our case, $x=x+y$ , and it is close to $0$... This is what I can't figure out. – homogenity May 14 '14 at 10:41

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