Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove that this function is surjective and monotonic?

$$f:\left]-5,5\right[\to \left[0,1\right], \ x\mapsto e^{-x^2}$$

share|improve this question
What is $f(-\pi)$? –  Hagen von Eitzen Feb 3 '13 at 15:33
Does there exist a preimage of 0? –  Abhra Abir Kundu Feb 3 '13 at 15:34
sorry the function is $x\rightarrow e^{-x^2}$ –  Devid Feb 3 '13 at 15:36
add comment

2 Answers

up vote 6 down vote accepted

First of all, it isn't monotonic (as it is an even, non-constant function). Second of all, at no point is $f(x)=0$, so it wouldn't be surjective as stated.

share|improve this answer
sorry the function is $x\rightarrow e^{-x^2}$ –  Devid Feb 3 '13 at 15:37
That takes care of a few problems, but not all of them. See my edit. –  Cameron Buie Feb 3 '13 at 15:40
Ok but you proved that the function is not surjective because $f(x)\neq 0$. But how do you know that it is not monotonic, how do you prove that ? –  Devid Feb 3 '13 at 15:47
Note that $f(0)=1$, then take any $0<x<5$ and show that $f(-x)=f(x)<1$. Either that, or prove the more general result that no even, non-constant function on an open interval about $x=0$ can be monotonic. –  Cameron Buie Feb 3 '13 at 15:48
@Devid: Yes, that's certainly true. We generally only refer to a function as "monotonic" if it is increasing on its whole domain or decreasing on its whole domain. I see what you meant, now, though. –  Cameron Buie Feb 4 '13 at 2:34
show 6 more comments

There is no pre image of zero.

share|improve this answer
Try reformulating your answer so it doesn't look as a question. (And there is no need for eight question marks.) –  mrf Feb 3 '13 at 15:56
Thanks @mrf for this advice. –  Abhra Abir Kundu Feb 3 '13 at 16:05
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.