\beginsection 18.7

Prove that $x2^x=1$ for some $x$ in $(0,1)$.

\medskip

We define the function $f(x)=x2^x$.
For this function we have $f(0)=0$ and $f(1)=2$.
Since $f(0)<1<f(1)$, by IVT $f(x)=1$ must occur somewhere in the
closed interval $[0,1]$.
We have $f(0)\ne1$ and $f(1)\ne1$ so $x2^x=1$ must occur somewhere in the
open interval $(0,1)$.