Show (0,0) is a saddle point of the function 2x^3 + 6xy + 3y^2

Question

anonymous · Answer

My second derivative test failed since it equaled 0, so I am not certain how to proceed.

waynex · Answer

Can you show what you did for the second derivative test? My second derivative test shows "AC-B^2" to be negative, which is indeed indicative of a saddle point.

anonymous · Answer

$$12x\cdot6-[6(x^2+y)\cdot6(x+y)]^2=12\cdot0\cdot6-[6(0+0)\cdot6(0+0)]^2=0$$

waynex · Answer

It looks like you took did this:$$F _{xx}*F _{yy}-[F _{x}*F _{y}]^{2}.$$What you need to do is this:$$F _{xx}*F _{yy}-(F _{xy})^{2}.$$Notice that the last part there is the second derivative with respect to x and y. You took the first derivative with respect to x and multiplied by the first derivative with respect to y.