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?

  • 1
    I think you need to find the range of f(x,y), not y. You need to think about what all values f(x,y) can take for any real x, y – Neo Oct 11 at 9:33
up vote 5 down vote accepted

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

  • Why does it need to be less than 1 though? - Thanks for the edit! – Hews Oct 11 at 9:36
  • 1
    @Hews I added something. See now. – Michael Rozenberg Oct 11 at 9:38

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: $$0<f(t)\le 1 \iff 0<f(x,y)\le 1.$$

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