I know the domain is $\mathbb{R^2}$.
Is the range $\mathbb{R}$?
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1The output is a scalar, how could the range even be $\mathbb R^2$? – Git Gud Jul 07 '14 at 22:47
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$f:\mathbb{R}^2 \rightarrow \mathbb{R}$. – Kaj Hansen Jul 07 '14 at 22:50
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If you raise $e$ to a power that is a real number, you always get a positive number. So $0$ and negative numbers cannot be within the image of this function. – Michael Hardy Jul 08 '14 at 00:03
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$$x^2+y^2\ge 0 \implies e^{-(x^2+y^2)}\le e^{-0}=1$$
$$x^2+y^2<\infty \implies e^{-(x^2+y^2)}\gt e^{-\infty}=0$$
So the range of $f(x,y)$ is $(0,1]$.
mike
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1Note that this answer just proves that $\text{ran}(f)\subseteq ]0,1]$. You still need to prove the other inclusion. – Git Gud Jul 07 '14 at 23:05
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What are the possible values? For example, $f(1,0)=e^{-1}$, so $e^{-1}$ is in the range. Note $e^{-1}$ is not in $\mathbb{R}^2$.
Empy2
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@RoyM. Now what is special about $e^2$? What other numbers can't you get? – Git Gud Jul 07 '14 at 23:04
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You are looking for all possible values of this function.
First note that $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, so the possible values of $f$ are real number.
Next notice that $x^2+y^2$ for $(x,y) \in \mathbb{R}^2$ can reach all non-negative real value (choose real non-negative $t \in \mathbb{R}$, $x^2+y^2$ reach value $t$ for $(x,y)=(\sqrt{t},0)$ . Note that for $(x,y) \in\mathbb{R}^2$ $x^2+y^2$ don't reach negative value.
Now you can check range of $e^{-x^2-y^2}$:you know that range of $x^2+y^2$ (all non-negative real numbers), so $e^{-x^2-y^2}$ reach the same values like $e^{-t}$ for non-negative real $t$, so the range is $(0,1]$.
agha
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