Let us consider a function $f \in \mathcal{C}^{2n+1}(\mathbb{R})$ and a fixed $\varepsilon$>0. We say that a function $\phi \in \mathcal{C}^{2n+1}(\mathbb{R})$ is an $\varepsilon-$interpolating function of $f$ if: \begin{cases} \phi^{(k)}(0) &= f^{(k)}(0) + A_k \varepsilon^{2n+1-k},\quad k=0, \cdots, n-1,\\ \phi^{(k)}(\varepsilon) &= f^{(k)}(\varepsilon)+B_k \varepsilon^{2n+1-k},\quad k=0, \cdots, n-1, \end{cases} where the constants $A_{k},B_{k}$ do not depend on $\varepsilon$. In other words, $\phi$ approximates $f$ and its derivatives at the extremes of $[0,\varepsilon]$ with error proportional to some power of $\varepsilon$. Besides that, we say that $\phi$ is $\varepsilon-$integral preserving to $f$ if \begin{equation} \int_{0}^{\varepsilon} \phi(x) \,dx = \int_{0}^{\varepsilon} f(x) \,dx\\ \end{equation}
Question: If $\phi$ is an $\varepsilon-$interpolating function and $\varepsilon-$integral preserving to $f$, then how well $\phi(x)$ approximates $f(x)$ for $x \in ]0,\varepsilon[$? In particular, does exist a constant $C$ that do not depend on $\varepsilon$ such that: \begin{equation} |f(x)-\phi(x)| \leq C \varepsilon^{2n+1} \end{equation} I'm not sure if this is true, but my intuition says that this result may be true. Besides that, if it is helpful, I am willing to assume that $\phi$ is a polynomial of degree $\leq 2n$.
Attempt: For $x \in ]0, \varepsilon[$, I am considering the following Taylor's expansions: \begin{align} \phi(x) = \sum_{k=0}^{n-1}\phi^{(k)}(0)\frac{x^k}{k!} + \sum_{k=n}^{2n}\phi^{(k)}(0)\frac{x^{2n+1}}{k!}+ \phi^{(2n+1)}(\alpha)\frac{x^{2n+1}}{(2n+1)!}\\ f(x) = \sum_{k=0}^{n-1}f^{(k)}(0)\frac{x^k}{k!} + \sum_{k=n}^{2n}f^{(k)}(0)\frac{x^{2n+1}}{k!} + f^{(2n+1)}(\beta)\frac{x^{2n+1}}{(2n+1)!} \end{align} Subtracting these equations, we get \begin{align} f(x) -\phi(x) &= \sum_{k=0}^{n-1}\big(f^{(k)}(0)-\phi^{(k)}(0)\big)\frac{x^k}{k!} + \sum_{k=n}^{2n}\big(f^{(k)}(0)-\phi^{(k)}(0)\big)\frac{x^k}{k!}\\ &= \sum_{k=0}^{n-1}{A_k \frac{x^k}{k!} \varepsilon^{2n+1-k}} \quad +\quad \sum_{k=n}^{2n}\big(f^{(k)}(0)-\phi^{(k)}(0)\big)\frac{x^k}{k!} + \bigg(f^{(2n+1)}(\beta)-\phi^{(2n+1)}(\alpha)\bigg)\frac{x^{2n+1}}{(2n+1)!} \end{align} Therefore: \begin{align} |f(x) -\phi(x)| &\leq \varepsilon^{2n+1}\sum_{k=0}^{n-1}\frac{|A_k|}{k!} + \sum_{k=n}^{2n}\big|f^{(k)}(0)-\phi^{(k)}(0)\big|\frac{\varepsilon^k}{k!} \\ &+\bigg|f^{(2n+1)}(\beta)-\phi^{(2n+1)}(\alpha)\bigg|\frac{\varepsilon^{2n+1}}{(2n+1)!} \end{align}
It remains to bound the term $\big|f^{(k)}(0)-\phi^{(k)}(0)\big|$ for $k=n, \cdots, 2n$. If we prove that \begin{equation} \big|f^{(k)}(0)-\phi^{(k)}(0)\big| \leq A_k \varepsilon^{2n+1-k} \quad \text{for} \quad k=n, \cdots, 2n \end{equation} for constants $A_k$ that do not depend on $\varepsilon$, then we are done.
So far, I have not used the information $\phi^{(k)}(\varepsilon) = f^{(k)}(\varepsilon)+B_k \varepsilon^{2n+1-k}\quad$ for $k=0, \cdots, n-1$ and I do not see how this helps to bound $\big|f^{(k)}(0)-\phi^{(k)}(0)\big|\quad$ for $\quad k=n, \cdots, 2n$.
I do appreciate any help. Thanks in advance.