Lagrange Error Value for $f(x)=\frac{1}{x}$ I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= \frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$
Step $1$ is to calculate the Taylor-Polynomial:
$$T_3(x,5)= \frac{1}{5}-\frac{(x-5)}{25}-\frac{(x-5)^2}{125}-\frac{(x-5)^3}{625}$$
I then used the Remainder formula $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!} \cdot (x-a)^{(n+1)}$. From this I get:
$$R_3(x)=\frac{f^{(4)}(\xi)}{4!} \cdot (x-5)^4 =\frac{-1}{\xi^4} \cdot (x-5)^4 = \frac{-(x-5)^4}{\xi^4}$$
Here's where I'm not sure how to proceed. I thought that I could make $2$ cases to try to find the Worst-Case scenario by checking which value of $\xi_1=4, \xi_2=6$ gives the Worst-Case scenario btu then I'm still left with either 
$$R_{n_1}(x) =\frac{-(x-5)^4}{4^4} \text{ or } R_{n_2}(x) =\frac{-(x-5)^4}{6^4}$$
but in either case I'm still left without an exact value for the error, which makes my answer incorrect. 
How can I obtain an exact value for the error of the Taylor-Polynomial in the interval $x \in [4,6]$?
 A: Consider the function
$$f(x) = \dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$\begin{array}{|c|c|}
\hline \text{Derivatives} & \text{Values at x} =5 \\\hline
  f(x) = \dfrac{1}{x} & \dfrac{1}{5} \\\hline
  f'(x) = -\dfrac{1}{x^2} & -\dfrac{1}{25} \\\hline
  f''(x)=\dfrac{2}{x^3} & \dfrac{2}{125} \\\hline
  f^{(3)}(x) = -\dfrac{6}{x^4} & -\dfrac{6}{625} \\\hline
f^{(4)}(x) = \dfrac{24}{x^5} & \dfrac{24}{3125} \\\hline
 \end{array}$$
So the third degree Taylor polynomial is
$$ T_3(x,5)= \frac{1}{5}-\frac{(x-5)}{25}-\frac{(x-5)^2}{125}-\frac{(x-5)^3}{625}$$
The Lagrange remainder is given by
$$\displaystyle \left|R_n(x)\right|=\left|\dfrac{f^{(n+1)}(\xi)}{(n+1)!}\right| \cdot \left|(x-a)^{(n+1)}\right|$$
Find the Lagrange Remainder centered at $a = 5$
$$|R_3(x)|=\left|\dfrac{f^{(4)}(\xi)}{(4)!}\right| \cdot \left|(x-5)^{4}\right| = \left|\dfrac{1}{\xi^5}\right| \cdot \left|(x-5)^{4}\right|$$
On the interval $4 \le \xi \le 6$, the maximum of $\left|\dfrac{1}{\xi^5}\right|$ is at $\xi = 4$, hence, the maximum error is given by 
$$|R_3(x)| =  \left|\dfrac{f^{(4)}(\xi)}{(4)!}\right|  \cdot \left|(x-5)^{4}\right| = \dfrac{1}{24576} \cdot \max \left|(x-5)^{4}\right|$$
Note that we can also find the maximum of that last expression as $1$ at $x = 4$.
