Find range of exponent

Question

I have the following function I need to find the range for and I'm not sure if I'm on the right direction.

$f(x,y) = e^{-x^2-(y-1)^2}$

$x$ & $y$ are real-numbers.

I'm thinking that the range is "all real values for $y$ that are $> 0$."

Is this right?

Answer 1

Since $-x^2-(y-1)^2\leq0$ and $g(x)=e^x$ increases, we obtain: $$0<e^{-x^2-(y-1)^2}\leq e^0=1.$$

Answer 2

Note that: $$f(x,y) = e^{-x^2-(y-1)^2}=e^{-(x^2+(y-1)^2)},(x,y)\in \mathbb R^2 \iff \\ f(t)=e^{-t}, t\ge 0.$$ Since $f(0)=1$; $f'(t)=-e^{-t}<0, t>0$ and $\lim_\limits{t\to +\infty}f(t)=0$, we get: $$0<f(t)\le 1 \iff 0<f(x,y)\le 1.$$