I am having a problem with the following equation $$e^{2x}(2x−1)+1=0$$
By trial and error, you can see that the root is $x=0$, but how can I actually find it?
$$e^{2x}=\frac{-1}{2x-1}$$
As we can see, $e^{2x} > 0, \forall x \in \Re$, but $\frac{-1}{2x-1} > 0 , x \in (-\infty,\frac{1}{2})$, and $\frac{-1}{2x-1}<0, x\in(\frac{1}{2},\infty)$.
When the $x\to-\infty$, LHS is decreasing, and RHS is increasing, but when $x\to\infty$ LHS is increasing, and RHS is decreasing. How can I find the root of this equation? It is clear that it has a root, because one can easily find two points where in one function is positive, and in the other one it is negative.
To solve $$e^{2x}(2x-1)+1=0$$ you will need to use the Lambert-W Function. You can read that article about it if you want to, but, to make a long story short, it is the inverse of $$y=xe^x$$ so that $$W(xe^x)=x$$ and $$W(x)e^{W(x)}=x$$ Using this info, you can transform your equation into $$e^{2x}(1-2x)=1$$ Letting $k=1-2x$: $$ke^{1-k}=1$$ $$-ke^{1-k}=-1$$ $$-ke^{-k}=\frac{1}{e}$$ and so, applying the Lambert $W$ function, $$-k=W\bigg(-\frac{1}{e}\bigg)$$ and $$2x-1=W\bigg(-\frac{1}{e}\bigg)$$ This value of the Lambert-W function is known to be (and, upon inspection, is obvious to be) $-1$, so we have $$2x-1=-1$$ $$x=0$$ Does that answer your question? If not, please clarify what you meant and I will be happy to change my answer.