$e^{x-1}$ has only one fixed point

Question

How does one show that the function $E(x)=e^{x-1}$ has only one fixed point? We know that there is only one integer solution for $e^{1-x} = x$, which is $x=1$, but Wolfram Alpha also gives a second answer in terms of some analytic function $W_n$, which I don't know.

How can it be proved that there is indeed one fixed point for $E(x)$?

We know that $E(x) \ge x$ for all $x$. One way is to prove that the equality holds only for $x=1$.

Answer 1

Let $$D(x) = E(x) -x = e^{x-1}-x$$ then you need to find all places where $D(x) = 0$. Note that $$D'(x) = e^{x-1}-1$$ and $$D''(x) = e^{x-1} > 0,$$ so $D$ is always concave up.

Moreover, $D'(x) > 0$ iff $x>1$, $D'(x) < 0$ for $x<1$. So $D(x)=0$ at $x=1$, and to the right one $1$, $D$ is increasing, and to the left - decreasing, so it never hits 0 again.

Answer 2

You want points for which $e^{x-1} = x$. If you plot those two functions ($y = e^{x-1}$ and $y = x$), you will see that they meet, at a tangent, at only one point: $(1, 1)$.

Answer 3

Look at the function $e^{x-1}-x$. You may differentiate it to show that prior to the zero at $x=1$, it is only decreasing, and after that zero, it is only increasing. Therefore $x=1$ must be the only solution.