Functions of Several Variables — Question 6

Question 6 – Solution

Strategy Compare $|x^2-y^2|$ with $x^2+y^2$, then solve the level equation.

Step 1: Bounds Since $x^2,y^2\ge 0$, $$|x^2-y^2|\le x^2+y^2.$$ The denominator is positive off the origin, so $-1\le f\le 1$.

Step 2: Endpoints and intermediate values On the $x$-axis, $f=1$; on the $y$-axis, $f=-1$. For $-1<c<1$, $$(1-c)x^2=(1+c)y^2.$$ Choosing $y=1$ and $x=\sqrt{(1+c)/(1-c)}$ realizes $c$. Hence $$\boxed{\operatorname{range}(f)=[-1,1]}.$$

Step 3: Levels For $-1<c<1$, the level set is the pair of lines $$\boxed{y=\pm\sqrt{\frac{1-c}{1+c}}\,x},$$ with the origin removed. At $c=1$ and $c=-1$, the levels are the punctured $x$- and $y$-axes.