Let $f:[2,7] \rightarrow \Bbb{R}$ be a continuous function and for given $\epsilon >0$,we have to prove that there exists a polynomial $p$ such that $f(2)=p(2)$, $p'(2) = 0$ and $\sup\{|p(x) - f(x)|\}<\epsilon$.
I think that this follows from the Weierstrass approximation theorem ($p$ approximates $f$ uniformly) but how to think of the polynomial $p$ explicitly? Any ideas how to give an explicit $p$ or to show that there exists such a polynomial $p$?
I will show how to do this when the interval is $[0,1]$ and let you use a transformation to get the result for $[2,7]$. Given a continuous function $f$ on $[0,1]$ and $\epsilon >0$ we want to find a polynomial $p$ such that $p(0)=f(0), p'(0)=0$ and $|f(x)-p(x)| <\epsilon$ for all $x$. Let $g(x)=f(\sqrt x)$. The $g$ is a continuous function, so there exists a polynomial $q$ such that $|g(x)-q(x)| <\epsilon /2$ for all $x$. This gives $|f(x)-p_0(x)| <\epsilon /2$ for all $x$ where $p_0(x)=q(x^{2})$. Let $p(x)=p_0(x)+f(0)-p_0(0)$. Then $p$ is a polynomial, $p(0)=f(0)$ and since $p'(x)=2xq'(x^{2})$ we have $p'(0)=0$. Note that $|f(0)-p_0(0)| <\epsilon /2$. Hence $$|f(x)-p(x)|=|f(x)-p_0(x)-f(0)+p_0(0)|$$ $$\leq |f(x)-p_0(x)|+|f(0)-p_0(0)| <\epsilon /2+\epsilon /2=\epsilon $$.