find upper bound M taylor inequality Let $f(x) = \frac{1}{1-x}$
Find an upper bound M for $|f^{(n+1)}(x)|$ on the interval $[-1/2,1/2]$
I found derivative of the function and plugged in the points to see which one gives the largest value.
$f^{'}(x)=\frac{1}{(1-x)^2}$
$f^{'}(1/2)=4$
The solution was:  $2^{n+2}(n+1)!$
I know the formula $|R_n(x)|= \frac{M|x-a|^{n+1}}{(n+1)!}$ but not sure how they got the M. Any help?
 A: That is because you need to find the largest value of the $n$th derivative of $f$ over the interval [-0.5,0.5]. Given $f(x) = \dfrac{1}{1-x}$,
\begin{align*}
f'(x) & = \frac{1}{(1-x)^2}.\\
f''(x) & = \frac{1\cdot 2}{(1-x)^3}.\\
f'''(x) & = \frac{1\cdot 2\cdot 3}{(1-x)^4}.\\
\vdots & \qquad \vdots \\
f^{(n)}(x) & = \frac{1\cdot 2 \cdot\ldots\cdot n}{(1-x)^{n+1}} = \frac{n!}{(1-x)^{n+1}}.
\end{align*}
Thus, the maximum of $f^{(n)}(x)$ over the interval $[-0.5,0.5]$ is given by
$$ \max_{x\in [-0.5,0.5]} f^{(n)}(x) = \max_{x\in [-0.5,0.5]} \frac{n!}{(1-x)^{n+1}} = \frac{n!}{(1-0.5)^{n+1}} = 2^{n+1}n!.$$
A: Let $f(x)=\frac{1}{1-x}$.  Then, we have
$$\begin{align}
f^{(1)}(x)&=1\cdot(1-x)^{-2}\\\\
f^{(2)}(x)&=2\cdot1\cdot(1-x)^{-3}\\\\
f^{(3)}(x)&=3\cdot 2\cdot 1\cdot(1-x)^{-4}\\\\
&\,\,\,\vdots\\\\
f^{(n+1)}(x)&=(n+1)!\cdot(1-x)^{-{(n+2)}}\\\\
\end{align}$$ 
Inasmuch as $f^{(n+1)}(x)$ is monotonically increasing on $x<1$, then we can assert that
$$\max_{x\in[-1/2,1/2]}f^{(n+1)}(x)=f^{(n+1)}(1/2)=(n+1)!2^{n+2}$$
as was to be shown!
