Is every closed interval domain, continuous rational function equal to a polynomial? Given any 2 non zero polynomials with real coefficients $\mathbf Pn(x)$ and non-constant $\mathbf Pd(x)$ forming the rational function $\mathbf f(x)$ = $\mathbf Pn(x)$ / $\mathbf Pd(x)$.
 Restrict its domain $\mathbb R$ to a closed interval $\mathbb I$ such that $\mathbf f(x)$ is continuous over $\mathbb I$.

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*Is there for every such function $\mathbf f(x)$ a polynomial $\mathbf P(x)$ with real coefficients such that for any $\mathbf x$ $\in$ $\mathbb I$, $\mathbf f(x)$ = $\mathbf P(x)$?

*Is there for no non-trivial such functions $\mathbf f(x)$ a polynomial $\mathbf P(x)$ with real coefficient such that for any $\mathbf x$ $\in$ $\mathbb I$, $\mathbf f(x)$ = $\mathbf P(x)$??

Context: I'm working on interpolation/regression curves, that is I'm trying to identify families of curves with or without certain features.
Here I'm wondering if there's any loss of generality from working with polynomials VS rational functions when we consider only finite intervals and continuous curves: no trivial line of reasoning come to my mind.
 A: 

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*Is there for every such function $\mathbf f(x)$ a polynomial $\mathbf P(x)$ with real coefficients such that for any $\mathbf x$ $\in$ $\mathbb I$, $\mathbf f(x)$ = $\mathbf P(x)$?

*Is there for no non-trivial such functions $\mathbf f(x)$ a polynomial $\mathbf P(x)$ with real coefficient such that for any $\mathbf x$ $\in$ $\mathbb I$, $\mathbf f(x)$ = $\mathbf P(x)$??


There does not exist such a polynomial, except for the trivial case when the rational function is itself a polynomial, which answers both questions above in the negative.
This follows from the stronger statement: if a real (or complex) rational function equals a polynomial at infinitely many points then the rational function is identically equal to the polynomial on $\,\mathbb R\,$ (or $\,\mathbb C\,$), and this can only happen when the denominator of the rational function divides the numerator as a polynomial, so the rational function reduces to the quotient polynomial after the division.
Let $\,f(x) = \frac{P_n(x)}{P_d(x)}\,$ be a rational function, and let $T$ be an infinite set of values such that $\,P_d(t) \ne 0\,$ and $\,f(t) = P(t)\,$ $\;\forall t \in T\,$ for some polynomial $\,P(x)\,$.
Let the polynomial $\,R(x) = P_d(x) \cdot P(x) - P_n(x)\,$, then for $\,\forall t \in T\,$, and given that $\,P_d(t) \ne 0\,$:
$$
R(t) = P_d(t) \cdot P(t) - P_n(t) = P_d(t) \cdot \left(P(t) - \frac{P_n(t)}{P_d(t)}\right) = P_d(t) \cdot \big(P(t) - f(t)\big) = 0
$$
Therefore, each $\,t \in T\,$ is a root of $\,R(x)\,$. But a polynomial of degree $\,n \ge 1\,$ has at most $\,n\,$ roots by the factor theorem and FTA (fundamental theorem of algebra) or, in other words, the only polynomial with infinitely many roots is the zero polynomial. It follows that $\,R(x)\,$ is the zero polynomial, so $\,P_d(x) \cdot P(x) - P_n(x) = 0$ $\iff\;P_n(x) = P_d(x) \cdot P(x)\,$. This proves that $\,P_d(x) \mid P_n(x)\,$, so the rational function $\,\frac{P_n(x)}{P_d(x)}\,$ simplifies to a polynomial.
