Stationary points for $f(x,y)=(x^2+y^2-4)^2-x^2$

Question

Find extreme values for $$f(x,y)=(x^2+y^2-4)^2-x^2$$

My attempt: I calculated $$f_x=4x^3+4xy^2-18x=0$$ $$=2x(2x^2+2y^2-9)=0$$ and $$f_y=(2y)(x^2+y^2-4-x)+(x^2+y^2-4+x)(2y)$$ $$4y(x^2+y^2-4)=0$$

So,

$y=0$ or $x^2+y^2=4$

I got stuck after this.How will I proceed?

Answer 1

Both the partial derivatives vanish when $$(x=0\vee x^2+y^2=\tfrac{9}{2})\wedge(y=0\vee x^2+y^2=4)$$ hence the stationary points are $(0,0),(0,\pm 2),(\pm\frac{3}{\sqrt{2}},0)$. Evaluate $f$ at these points to get the absolute minimum. There is no absolute maximum since your function is clearly unbounded on the line $x=0$.