Given a function $f(x) = \ln(x+1)$ and its inverse $f^{-1}(x) = e^x - 1$ , find the invariant points.
After equating the two $y$ values of a function, $f(x) = f^{-1}(x)$, I get:
$e^x - 1 = \ln(x+1)$
I am unable to solve this equation since I don't know whether I should just express the $e$ value or if there is a more sophisticated approach?
Saying $f(x) = f^{-1}(x)$ means that they intersect along the line $y=x$. We could equivalently ask for either
$$x = e^x-1$$
$$x = \ln(x+1)$$
Which both have solutions at $x=0$ by guessing. A little IVT magic proves that this is the only solution for either equation.