I am doing critical points and using the second derivative test (multivariable version)
1. The problem statement, all variables and given/known data
[tex]f(x,y) = (x^2+y^2)e^{x^2-y^2}[/tex]
Issue I am having is with the system of equations to get the critical points from partial wrt x, wrt y
3. The attempt at a solution
[tex]f_{x} = 2xe^{x^2-y^2}+2xe^{x^2-y^2}(x^2+y^2)[/tex]
[tex]f_{y} = 2ye^{x^2-y^2}-2ye^{x^2-y^2}(x^2+y^2)[/tex]
It is pretty easy to see (0,0) makes both equal to zero
I know +/- 1 = y and x = 0 is a solution as well.
[tex](x^2+y^2)[/tex] is never negative let alone zero [over the reals]
[tex]e^{x^2-y^2}[/tex] won't be zero, it will get infinitesimally close to zero for values of square of y > square of x
and 2x+2x=0 only when x = 0
and 2y-2y=0 for all real values of y
how do I arrive at y = +/- 1 [ I can see it in the line just above but then I can use any real value for y]
1. The problem statement, all variables and given/known data
[tex]f(x,y) = (x^2+y^2)e^{x^2-y^2}[/tex]
Issue I am having is with the system of equations to get the critical points from partial wrt x, wrt y
3. The attempt at a solution
[tex]f_{x} = 2xe^{x^2-y^2}+2xe^{x^2-y^2}(x^2+y^2)[/tex]
[tex]f_{y} = 2ye^{x^2-y^2}-2ye^{x^2-y^2}(x^2+y^2)[/tex]
It is pretty easy to see (0,0) makes both equal to zero
I know +/- 1 = y and x = 0 is a solution as well.
[tex](x^2+y^2)[/tex] is never negative let alone zero [over the reals]
[tex]e^{x^2-y^2}[/tex] won't be zero, it will get infinitesimally close to zero for values of square of y > square of x
and 2x+2x=0 only when x = 0
and 2y-2y=0 for all real values of y
how do I arrive at y = +/- 1 [ I can see it in the line just above but then I can use any real value for y]
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