Suppose $f:\mathbb{R}\to\mathbb{R}$ has n continuous derivatives. Show that for every $x_0\in\mathbb{R}$, there exist polynomials $P$& $Q$ of degree $n$ and an $\epsilon>0$ s.t. $P(x)\leq{f(x)}\leq{Q(x)}$ for all $x\in[x_0,x_0+\epsilon]$ and $P(x)-Q(x)=\lambda(x-x_0)^n$ for some $\lambda\geq{0}$.
I think I have the start of a proof for the theorem but I'm not quite sure.
My Proof: Expand $f(x)$ into a Taylor polynomial about $x_0$ to get $$f(x)=\sum_{k=0}^{n-1}{\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k}+R_{n}^{x_0}(x)=P_{n-1}^{x_0}(x)+R_{n}^{x_0}(x)$$ where $R_n^{x_0}(x)=\frac{f^{(n)}(c)}{n!}(x-x_0)^n$ is the remainder term in the Lagrangian form, where $c$ is some number in between $x_0$& $x$. If $f(c)\geq{0}$ define $$P(x)=P_{n-1}^{x_0}(x)+\frac{f^{(n)}(c)}{(n+1)!}(x-x_0)^n$$ and $$Q(x)=P_{n-1}^{x_0}(x)+\frac{f^{(n)}(c)}{(n-1)!}(x-x_0)^n.$$ For $f(c)<0$ we can do a similar case. Now because $(n-1)!\leq{n!}\leq{(n+1)!}$ we have $P(x)\leq{f(x)}\leq{Q(x)}.$ From here I don't really know what to do, I've proved the theorem for a point $x$ but to prove it for the interval maybe I could take a max of $f^{(n)}(x)$ on an interval?
