Check whether the function #f:RR^2->RR# defined by #f(x,y)=2x^4-3x^2y+y^2# has an extrema at #(0,0)#?

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Answer 1 · 2018-04-14T23:13:59

No - #(0, 0)# is a saddle point.

Given:

#f(x, y) = 2x^4-3x^2y+y^2#

Note that:

#del/(del x) f(x, y) = 8x^3-6xy#

#del/(del y) f(x, y) = -3x^2+2y#

So both partial derivatives are #0# at #(0,0)#. So #f(x, y)# has some sort of critical point there.

We can factor #f(x, y)# as:

#2x^4-3x^2y+y^2 = (x^2-y)(2x^2-y)#

Hence approaching #(0, 0)# along either of the curves #y = x^2# or #y = 2x^2#, the function is constantly #0#.

If we approach #(0, 0)# along the curve #y = 3/2 x^2#, then #f(x, y) < 0#

If we approach #(0, 0)# along the line #y = 0# or #x = 0# then #f(x, y) > 0#.

So #(0, 0)# is some kind of saddle point, i.e. not a local maximum or minimum.