Infimum and supremum of two variable function

Question

How can I find the infimum and supremum in $\mathbb{R}^{2} $ of this function $$ f(x,y)=(2x^2+y^2-1)(x^2+y^2-1)+1 $$? Thanks

EDIT:

Forgive me if I did not add my thoughts but I did not know where to start .

Answer 1

The function isn't bounded above since

$$f(x,0)=2x^4-3x^2+2\rightarrow\infty$$

Since the function is even in both variables we can work only with $\;x,y\ge0\;$ , and

$$f(x,y)\xrightarrow[(x,y)\to(0,0)]{} 1+1=2$$

But for any point on the circle $\;x^2+y^2=1\;$ or on the ellipse $\;2x^2+y^2=1\;$ the value of the function is $\;1\;$ : $\;f(x,y)= 1$

Yet things get more interesting if we have values $\;(x,y)\;$ s.t. $\;x^2+y^2<1\;$ but $\;2x^2+y^2>1\;$, since then $\;f(x,y)<1\;$ . For example, with

$$\;f(0.9,0.1)=(1.62+0.01-1)(0.81+0.01-1)+1\cong0.89$$

Thus, this seems to depend on the parts $\;2x^2+y^2\;,\;\;x^2+y^2\;$ of the function, and since the former encloses the latter, we want points inside the disk $\;x^2+y^2<1\;$ but outside the ellipse $\;2x^2+y^2\le1\;$ .

The greatest difference between the values of points of this kind is on the main (horizontal) axis of the ellipse, so $\;y=0\;,\;\;\frac1{\sqrt2}\le x\le1\;$ , and we want the minimum value on this range of

$$f(x,0)=2x^4-3x^2+2\,.\;\;f\left(\frac1{\sqrt2},0\right)=\frac12-\frac32+2=1=f(1,0)\;,\;\;\text{and differentiating:}$$

$$f_x(x,0)=8x^3-6x=2x(4x^2-3)=0\iff x=\pm\frac{\sqrt3}2\;\;\text{(in our range)}$$

Differentiating twice it is easy to check this is a minimum point, and its value is

$$f\left(\frac{\sqrt3}2,0\right)=2\cdot\frac9{16}-3\cdot\frac34+2=\frac{14}{16}=\frac78$$