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 solve this for $x$?

$xe^x=-2/a$ with $(a \in \mathbb{R_0^+})$

$a$ can be any strict positive real number.

I need this because I'm searching for the root of a function to sketch a graph.

share|improve this question
5  
You can't do it with the usual elementary functions. But see the Wikipedia page on the Lambert $W$-function, especially the section on applications (J.M. or someone else will certainly jump in with some better links shortly). Concerning your notation: $(a \in \mathbb{R_0^+})$ means $a \gt 0$? –  t.b. Aug 24 '11 at 9:34
    
@Theo: I mean a has to be a strictly positive real number, so yes. However, I had to find this during my exam today, which is only in real numbers.. the link you provided uses complex numbers. –  Mats Aug 24 '11 at 9:39
1  
Strange... I would read that as $a \geq 0$. Anyway: Concerning your task in the exam: did you have to solve for $x$ or did you have to show that a solution $x$ exists? The latter can be achieved using the intermediate value theorem. –  t.b. Aug 24 '11 at 9:41
1  
I see. I'm sure somebody will help you with that. Unfortunately I don't have time right now. You may want to change your question in view of your last comment, because answerers will undoubtedly work something out using Lambert $W$ the way you ask it. If you want to know how to sketch the graph, in your last comment, you should say so. (notice the sketch which doesn't ask you to determine the root in the denominator exactly). –  t.b. Aug 24 '11 at 9:52
3  
Mats, the reals are a subset of the complex numbers so there's no issue. You can see what $W$ looks like on $\mathbb{R}$ on the graphic at the top of the linked Wikipedia page. The equation $$xe^x = b$$ only has real solution(s) if $b\ge -e^{-1}$. Also, nobody takes the chat warning seriously. :) –  anon Aug 24 '11 at 10:33
show 4 more comments

1 Answer 1

up vote 3 down vote accepted

I suspect anon's comment is what your exam paper was looking for: for $a$ sufficiently small, $ax e^x + 2 > 0$ always, so for a sketch of the graph you have something smooth with certain asymptotic behaviour.

For $a = 2e$, there is exactly one root, and you should show that you get blow-up of the same sign when you approach the singular point from the left and when you approach the singular point from the right. (Informally: $\lim_{x\to -1^+} (axe^x + 2)^{-1} = \lim_{x\to -1^-} (axe^x + 2)^{-1} = +\infty$)

For $a > 2e$, there are two roots, and at those two singularities, the left limit and the right limit have opposite signs, so you need to show that on the graph. In other words, I suspect the exam paper was asking for a qualitative depiction of the graph of the function, and not for a strictly quantitative depiction.

share|improve this answer
add comment

Your Answer

 
discard

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.