Monotone Increasing Function w/ an Upper Bound $\implies$ Unique Solution? Say I've shown that $f(x)$ is monotonically increasing on $(0, \infty)$ (by showing its derivative is always positive when $x > 0$). Say too I consider $F = \{\alpha\mid f(\alpha) \leq C \}$ so that $C$ is the least upper bound of the image of $F$. Is it reasonable to conclude that there is a unique $\alpha$ such that $f(\alpha) = C$? Or does this depend entirely on the $f(x)$ being considered?
For some possibly-unhelpful context, I feel an uncomfortable divide between convergence and equality (the unique doesn't help either). Part of me is absolutely sure this is a reasonable claim and another part of me screams madly against it—I am leaning towards the latter part. There must be unity or else I will pop.

No,
because $f$
may have $C$ as an asymptote.
Consider
$\tanh(x)
=\cfrac{\sinh(x)}{\cosh(x)}
=\cfrac{e^x-e^{-x}}{e^x+e^{-x}}
=\cfrac{e^{2x}-1}{e^{2x}+1}
=1-\cfrac{2}{e^{2x}+1}
$
which approaches $1$ 
but never reaches it.
