Positive solution for $e^x-x=c$

Question

In my problem I get the equation

(1):

$e^{x}-x=c$

with $c > 1$.

With my knowledge the analytical solution will be

(2):

$x = -W(-e^{-c})-c$.

With $c = 5.5555$, and this is my problem I only find the approximate solution:

$x = -5.55112$

Though, I know there is a second solution of approximately:

$x = 2.02559$.

So, this problem having two real solutions one negative and one positive, I wonder how to get the positive solution, because I am only interested in the postive solution $x > 0$.

Can you help me?

Answer 1

Consider the function $$f(x)=e^x-x-c\qquad \implies \qquad f'(x)=e^x-1$$ So, the first derivative cancels if $x=0$ which is a minimum by the second derivative test. Since $f(0)=-c <0$, you have two roots corresponding to the two branches of Lambert function.