Condition implying that the argmax of a function is piecewise continuous.

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Let $f_i$ be a smooth function mapping $[0,1)$ to $\Re$, for $i = 1, ... n$, and let $\bar f=\max f_i$. Define $T_i = \{ t \in [ 0 , 1 ) s.t. f_i(t) = \bar f(t) \}$. I'm looking for a condition on the $f_i$'s that implies the following property: for each $i$, $T_i$ is the union of finitely many intervals. Here's an example of the kind of thing I need to rule out: $f_1(t) = 1 - t$ and $f_2(t) = f_1(t) + c t^2 sin ( 1/t )$, where $c\in \Re$ can be chosen to be arbitrarily small. That is, $f_2$ wiggles around $f_1$ and, as $t$ approaches zero, the amplitude of the wiggles approaches zero, while the periodicity/frequency (??) of the wiggles increases without bound. Thus there are infinitely many sub-intervals of $T_1$ and infinitely many of $T_2$.

Can anybody give me a condition guaranteeing my property? I suspect that $f_2$ may not be analytic, but I don't know much about analytic functions, and not enough to prove it isn't. If $f_2$ is indeed not analytic, might analyticity be the condition that I need? I think this is wildly optimistic, but ...