Let us consider a polynomial function $P:[x_{1},x_{2}]\to[0,1]$ of degree $n\geq 2$. I would like to know if it is always possible to prove that it can be split into restrictions $\left.P\right|_{D_{k}}$, where $D_{k} = [d_{k},d_{k+1}]$ and $x_{1} = d_{1} < d_{2} < \ldots < d_{m+1} = x_{2}$, and the number $m$ is the least number of divisions necessary to ensure that $P_{k} := \left.P\right|_{D_{k}}$ is bijective.
If this is possible, since $P_{k}$ is continuously differentiable, the inverse function theorem ensures us that $P^{-1}_{k}$ is differentiable at $P_{k}(x_{0})$ if $P'_{k}(x_{0})\neq 0$. My argument is based on the following statement:
Inverse function theorem
Let $f:X\to Y$ be an invertible function, with inverse $f^{-1}:Y\to X$. Suppose that $x_{0}\in X$ and $y_{0}\in Y$ are such that $f(x_{0}) = y_{0}$. If $f$ is differentiable at $x_{0}$, $f^{-1}$ is continuous at $y_{0}$, and $f'(x_{0})\neq 0$, then $f^{-1}$ is differentiable at $y_{0}$ and \begin{align*} (f^{-1})'(f(x_{0})) = [f'(x_{0})]^{-1}. \end{align*}
Since $P_{k}$ is bijective and continuous, it is also monotonic. Thus we conclude that $P^{-1}_{k}$ is also continuous and monotonic. Based on my guess, it also seems true that $(P^{-1}_{k})'$ is continuous. My second question is: can we prove that $(P^{-1}_{k})'$ is continuous? If so, how do we do it?
Such questions are not homework. Any help is appreciated.
The derivative of a polynomial function is a polynomial function, and as such, it is a continuous function with only finitely many zeros. Those zeros divide the line into finitely many intervals. On the interior of each of those intervals, the derivative is either everywhere positive or everywhere negative. (That follows from the intermediate-value theorem, which says a continuous function never changes sign in an interval in which it is nowhere zero.)
If the derivative of a continuous function is everywhere positive or everywhere negative on an open interval, then the function is strictly monotone, and therefore bijective, on the corresponding closed interval.