Range of functions

Question

What is the range of the function: $$f(x,y)=\frac{7}{x^2+y^2+1}$$

I was thinking maybe, given that $x^2+y^2$ does not equal $-1$, then it could be any number from 0 to infinity? Or is that wrong?

Answer 1

Hint: It is $$x^2+y^2\geq 0$$ so $$x^2+y^2+1\geq 1$$

Answer 2

$x^2+y^2\ge0$.

$\therefore x^2+y^2+1\ge1$.

So, $f(x,y)=\dfrac 7{x^2+y^2+1}\le7$.

On the other hand, we also have $f(x,y)\gt0$.

Thus the range is $(0,7]$.

Answer 3

Surely the definition field is all $\mathbb R^2$. Then for the codomine you can see that the function $g:\mathbb R \to \mathbb R$ so that $g(t)=\frac 7{1+t^2}$ is a continuous and differentiable function in all $\mathbb R$ and $g’(t)=-7\frac{2t}{1+t^2}$. So $g$ is maximal for $t=0\implies g(0)=7$ is the maximum value of the function. Next $\lim_{t\to \infty} g(t)=0$. These fact make us sure $g$ has codomine $(0,7]$. So, $\forall c\in (0,7]\; \exists t\in \mathbb R\mid\, \frac 7{1+t^2}=c$ then for this specific value of $t$ they exist infinitely many couples $(x,y)\in\mathbb R^2$ so that $x^2+y^2=t^2$ and so codomine of $f$ contains $(0,7]$. Finally if it would exists $x,y\in\mathbb R$ so that $f(x,y)=c\in\mathbb R$ then it would exist $t\in\mathbb R$ so that $t^2=x^2+y^2$. We can conclude that $Im(f)=(0,7]$