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**superdude** show that the function $\displaystyle f(x,y)=xy^2-x^2y$ has a unique critical point. what kind is it?

I tried using the second derivative test. So from first derivatives I got the critical points (0,0) (1,0) and (1,2)

How? $\displaystyle f_x=y^2-2xy=y(y-2x)=0\,,\,\,f_y=2xy-x^2=x(2y-x)=0\,\Longrightarrow\,x=y=0$ $\displaystyle \Longrightarrow\,(0,0)$ is the only solution to this system, as can be checked...or better: check that $\displaystyle (1,0)\,,\,(1,2)$ are not zeroes of this sytem but $\displaystyle (,0) is.$[tex]

I then tried using the determinant where $\displaystyle D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$ but it was always inconclusive. For example D=0 for the point (0,0) and $\displaystyle f_{xx}=0$ for (1,0) and I don't know how to interpret that.

And you don't need to: critical points are points where the first order partial derivatives vanish, and that's what's been done above

I haven't learned Lagrange multipliers yet.