Determine the number of roots of $f(x)=xe^x-1 $ How can I find the number of roots of $f(x)=xe^x-1$? Is this problem solvable using Bolzano's and Rolle's theorems?
 A: Yes, you can use Rolle's theorem. After convincing yourself that a root exists, and every root must be positive, note that
$$f'(x) = x e^x + e^x$$
is strictly positive for $x > -1$; in fact, $f'(x) \ge 1$ for all $x \ge 0$.  Can you finish it from here with Rolle's theorem?
A: You are looking for values of $x$ where the graphs of $y=e^x$ and $y=\frac{1}{x}$ meet.
The first function is strictly increasing and positive. The second function is positive and strictly decreasing for positive $x$. Hence there is only one positive root.
A: Here is a more rigorous treatment:
The function is differentiable. Compute the derivative and look at its sign. The derivative is positive for x>-1 and negative for x<-1. Thus the function is increasing for x>-1 and decreasing for x<-1. 
Then plug in -1 in the function and look at the sign of f(-1). Plug in a "big" positive x and look at the sign. Compute the limit of the function for x-> minus infinity. 
Using the IVT (Intermediate value theorem) and the three values of the function you computed you should be able to prove the existence of one root.
Using the monotonicity of the function in the two intervals you prove uniqueness of the root.
