Find a polynomial which approximates $f(x) = \sqrt{x}$ in the interval $(4,5)$ within $10^{-8}$ 
Find a polynomial which approximates $f(x) = \sqrt{x}$ in the interval $(4,5)$ within $10^{-8}$.

I've been trying to solve this using Taylor's remainder theorem,
$$ |f(d) - P_n(d)| = \left| \frac{f^{n+1}(t)}{(n+1)!} \right| |d-c|^{n+1}$$
where $c,d \in (a,b)$ and $c < t < d$. However, I seem to be getting stuck finding the generalized derivative for the RHS. I keep getting
$$ \left| \frac{f^{n+1}(t)}{(n+1)!} \right| |d-c|^{n+1} = \frac{(-1)^n(2n-1)!}{2^{5n+4}(n-1)!(n+1)!}$$
What am I doing wrong?
 A: By the Lagrange form of the error in Taylor's Remainder Theorem, if $f^{(n+1)}(x)$ is continuous on an open interval $I$ which contains $a$ and $x\in I$, then there exists some $c$ between $a$ and $x$ such that
$$
R_n(x)={f^{(n+1)}(c)\over (n+1)!}(x-a)^{n+1}.
$$
My quick calculations yield $$f(x)=\sqrt{x}\implies f^{(n+1)}(x)={(-1)^n\over 2^{n+1}}(2n-1)!!\,x^{-(2n+1)/2}.$$ 
Taking $a=4.5$ (the midpoint of the given interval), we have $4\le x\le 5$ so $\displaystyle |x-4.5|^{n+1}\le {1\over 2^{n+1}}$, thus
\begin{align}
|R_n(x)|&=\left|{f^{(n+1)}(c)\over (n+1)!}(x-4.5)^{n+1}\right|\\
&={(2n-1)!!\over 2^{n+1}(n+1)!}{1\over c^{(2n+1)/2}}|x-4.5|^{n+1}\\
&\le {(2n-1)!!\over 2^{n+1}(n+1)!}\cdot{1\over 4^{(2n+1)/2}}\cdot{1\over 2^{n+1}}\\
&={(2n-1)!!\over (n+1)!}\cdot {1\over 2^{4n+3}}.
\end{align}
So you want to find the smallest positiver integer $n$ for which $$|R_n(x)|={(2n-1)!!\over (n+1)!}\cdot {1\over 2^{4n+3}}<10^{-8}$$ which occurs for $n=7$ (just write out that expression for the first few values of $n$). 
So take the $7$th degree Taylor polynomial for $f(x)=\sqrt{x}$ centered at $a=4.5$.
