TITLE: Prove $xe^x =2$ for some $x \in (0,1)$
QUESTION [6 upvotes]: We are trying to prove $xe^x =2$ for some $x \in (0,1)$. 

I know for certain that if this question were asking to prove the equality for some $x$ in the closed interval $[0,1]$ then I could apply the intermediate value theorem:

Let $f(x) = xe^x$ implying:

$f(0) = 0 \lt 2$ and $f(1) = 1\cdot e^1=e \gt 2$ and by the intermediate value theorem there must exist some $x \in [0,1]$ such that $f(x)=2$.

My question is if we can use the Intermediate Value Theorem in my original question - where we have $x \in (0,1)$ the open interval.

I know the discrepancy arises since the interval is open and there is no explicit definition of the function at the end-points $0$ and $1$.
Thank you guys!

REPLY [0 votes]: we have $f(0)=0$ and $f(1)=e>2$ then the function must have a root between $0$ and $1$.
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