Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is something not 0, then the level curve becomes a circle. And for big values of level curves, the circle disappears. Is there something I can use to find the range of this function?
 A: Take partial derivative of $f(x,y)$ and solve the system of $$\frac {\partial f}{\partial x}=0\implies x^2-y^2-2xy=1$$
$$\frac {\partial f}{\partial y}=0 \implies y^2-x^2-2xy=1$$ to find your local minimum and local maximum.
I found $$x=-y=\pm\frac {\sqrt 2 }{2}$$
Note that your function approaches zero for large values of $x$ and $y$
Thus the range is $$[-\sqrt 2/2, \sqrt 2/2]$$
A: Hint: Turn to polar coordinates with
$$\begin{cases}x&=&r \cos \theta \\ y&=&r \sin \theta \end{cases}$$
Your function becomes:
$$F(r,\theta)=\dfrac{r}{1+r^2}(\cos(\theta)-\sin(\theta))=\underbrace{\dfrac{r}{1+r^2}}_{f(r)}\sqrt{2}\cos(\theta-\pi/4)$$
Therefore, one has to maximize/minimize separately in $r$ and in $\theta$... knowing that $f$ has its minimum in $(-1,-\tfrac12)$ and its maximum in $(1,\tfrac12)$ (by a separate study of function $f$).
I leave you the conclusion...
A: For an alternative approach...
Notice $f(1,1)=0$ so $0$ is in the range of $f$. Now if $C\neq 0$ then $$f(x,y)=C \iff\Big(x-\frac{1}{2C}\Big)^2+\Big(y+\frac{1}{2C}\Big)^2=\frac{1}{2C^2}-1$$ The above relation contains at least one point in the $(x,y)-$plane iff $C\in \big[-\frac{1}{\sqrt{2}},0)\cup (0,\frac{1}{\sqrt{2}}\big]$. Putting everything together we see the range is $\Big[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big]$.
A: $\frac{x-y}{1+x^2+y^2}-z = 0$
You can find $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$
Or you can find the direction of normal to the surface(z) and the normal must have direction ratios as $\langle0\hat{i}+0\hat{j}±1\hat{k} \rangle$
$\nabla f(x,y,z)=$$\frac{2x(x-y)-(1+x^2+y^2)}{(1+x^2+y^2)^2}\hat{i}+\frac{2y(x-y)+(1+x^2+y^2)}{(1+x^2+y^2)^2}\hat{j}-\hat{k}$
Let us put $y = -x$
$2x(2x)-(1+x^2+x^2) = 0$
Solving above:
{$x,y$}  ≡ {($2^{-1/2}, -2^{-1/2}$), $(-2^{-1/2},2^{-1/2} $)}
$f(2^{-1/2}, -2^{-1/2})$, $f(-2^{-1/2},2^{-1/2})$
implies the function has only two such maxima and minima $\frac{1}{\sqrt{2}}$ and $\frac{-1}{\sqrt{2}}$
As, the denominator dominates over the numerators (Quadratic in nature)
The range of function: [$\frac{-1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$]
A: Let $k\in\mathbb{R}$. It is enough to find the $(x,y)$ in the domain of $f$ (all $\mathbb{R}^2$) for which the equality is true
$$\frac{x-y}{1+x^2+y^2}=k$$
It is clear that the equality is true when $k=0$ (in this case $\frac{x-y}{1+x^2+y^2}=0\Longleftrightarrow x-y=0\Longleftrightarrow x=y$). Suppose that $ k\neq 0$. Then
\begin{align}
\frac{x-y}{1+x^2+y^2}=k &\Longleftrightarrow x-y= k(1+x^2+y^2)\\ &\Longleftrightarrow x-y=k+kx^2+ky^2\\&\Longleftrightarrow kx^2+ky^2-x+y=-k \\&\Longleftrightarrow (kx^2-x)+(ky^2+y)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}\right)+k\left(y^2+\frac{y}{k}\right)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}+\frac{1}{4k^2}-\frac{1}{4k^2}\right)+k\left(y^2+\frac{y}{k}+\frac{1}{4k^2}-\frac{1}{4k^2}\right)=-k\\&\Longleftrightarrow k\left(x^2-\frac{x}{k}+\frac{1}{4k^2}\right)-\frac{1}{4k}+k\left(y^2+\frac{y}{k}+\frac{1}{4k^2}\right)-\frac{1}{4k}=-k\\& \Longleftrightarrow k\left(x-\frac{1}{2k}\right)^2+k\left(y+\frac{1}{2k}\right)^2=-k+\frac{1}{4k}+\frac{1}{4k}\\& \Longleftrightarrow k\left(x-\frac{1}{2k}\right)^2+k\left(y+\frac{1}{2k}\right)^2=\frac{1-2k^2}{2k}\\& \Longleftrightarrow \left(x-\frac{1}{2k}\right)^2+\left(y+\frac{1}{2k}\right)^2=\frac{1-2k^2}{2k^2}
\end{align}
since the left side of this equality is the sum of two squares, it makes sense if and only if \begin{align}
\frac{1-2k^2}{2k^2}\geq 0 & \Longleftrightarrow 1-2k^2\geq 0 \\& \Longleftrightarrow  1\geq 2k^2 \\& \Longleftrightarrow \frac{1}{2}\geq k^2 \\& \Longleftrightarrow |k|\leq \frac{1}{\sqrt{2}}.
\end{align}
It is concluded that the range of $ f $ is $\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right].$
A: The denominator of the expression for the function $ \ z \ = \   f(x,y) \ = \  \frac{x-y}{1+x^2+y^2} \ $ has "four-fold" symmetry about the origin, which is "broken" by the "diagonal" (anti-)symmetry in the numerator.  So this function has the symmetry property $ \ f(x,-y) \ = \ -f(-x,y) \ \ $ and it is reasonable to expect that the extremal points of the function will be described by $ \ (\pm x \ , \ \mp x) \ \ . $
If we take "slices" $ \ x + y \ = \ c \ $ through the function surface, we obtain "cross-sectional" curves
$$  z \ \ = \ \  \frac{x \ - \ (c-x)}{1 \ + \ x^2 \ + \ (c-x)^2} \ \ = \ \ \frac{2x \ - \ c}{ 2x^2 \ - \ 2cx \ + \ (1 + c^2)} \ \ . $$
It is somewhat unappealing to differentiate this as it stands.  By completing-the-square in the denominator, we find that we can write
$$  z \ \ = \ \   \frac{2x \ - \ c}{ 2·\left(x^2 \ - \ cx \ + \ \frac{c^2}{4} \right) \ \ + \ (1 + c^2) \ - \ \frac{c^2}{2}} \ \ = \ \ \frac{2·\left(x \ - \ \frac{c}{2} \right)}{ 2·\left(x \ - \ \frac{c}{2} \right)^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } $$ $$ \rightarrow \ \ \frac{2·u}{ 2·u^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } \ \ . $$
It is clear that the denominator of the expression is (still) never equal to zero, so we can establish that the derivative of this transformed function is zero for $ \ 4·u^2 \  + \ 2·\left(1 + \frac{c^2}{2} \right) \ - \ 2u·4u \ = \ 0 $ $ \Rightarrow \ u^2 \ = \ \frac12·\left(1 + \frac{c^2}{2} \right) \ \ . $  This locates the extremal points on the cross-sectional curve, with the extremal values being
$$ z \ \ = \ \frac{\pm \ \sqrt2 \ · \ \sqrt{1 + \frac{c^2}{2}}}{ \left(1 + \frac{c^2}{2} \right) \  + \ \left(1 + \frac{c^2}{2} \right) } \ \ = \ \ \pm \frac{  \sqrt2}{2} \ · \ \frac{1}{\sqrt{1 + \frac{c^2}{2}}} \ \ = \ \  \frac{\pm \ 1}{\sqrt{2 \ + \ c^2 }} \ \ . $$
We see from this result that $ \ \lim_{|c| \ \rightarrow \ \pm \infty} \ \ z \ = \ 0 \ \ , $ as you observed.  The extremal values are greatest in absolute-value for $ \ c = 0 \ \ , $ confirming that the extremal points occur on the line $ \ x + y \ = \ 0 \ \ . $ The absolute maxima and minima for the function are thus $ \ \pm \frac{ 1}{\sqrt2} \ \ . $
[We did not need to know the locations of the extrema, but we use our relations at $ \ c = 0 \ $ to find $ \ u \ = \ x - \frac{0}{2} \ = \ x \ = \ \pm \frac{ 1}{\sqrt2} \ $ and $ \ y = \ 0 - x \ = \ \mp \frac{ 1}{\sqrt2} \ \ . \ ] $
