Lagrange remainder to approximate $3^{2.1}$ less than 0.1 How do I solve this problem:
Use the appropriate Taylor polynomial $P_n(x,c)$ to estimate $3^{2.1}$ with error less than $0.1$, given $\ln 3$ is about $1.099$.
I understand that the remainder formula is:
$$R_k(x,c)= \frac{f^{k+1}(x)}{(k+1)!}(x-c)^{k+1}$$
And I can solve these problems when the question tells you c=0. 
So for $f(x)=3^x$, $f^{(k+1)}(x)=(\ln3)^{k+1}(3^x)$
How do I chose a $c$?
 A: You get to choose $c$.  You can use $c=0$ if you want, in which case you need to find a $k$ such that $\frac {(\ln 3)^{k+1}3^{2.1}}{(k+1)!}2.1^{k+1}\lt 0.1$  You can just try increasing $k$ until it works.  You are expanding around zero here.
If you want a smaller $k$, you can expand around $c=2$.  Your function is then $f(x)=9\cdot3^{x-2}$ and you are looking for the value of $3^{0.1}$.  so now your error is bounded by $\frac {(\ln 3)^{k+1}3^{0.1}}{(k+1)!}0.1^{k+1}\lt 0.1$ which will require a smaller $k$
A: With $f(x) = 3^x = e^{(\ln 3)x}$, you ought to expand the Taylor series around $c = 2$ so that with $x = 2.1$, the quantity being raised to successively higher powers is small:  $x - c = 0.1$.
The Taylor coefficients are
$$
a_k = \frac{f^{(k)}(2)}{k!} = \frac{(\ln 3)^k 3^2}{k!}
$$
Now, the remainder is no larger than
$$
R_k(2.1, 2)= \frac{f^{k + 1}(2.1)}{(k + 1)!}(2.1 - 2)^{k + 1} = \frac{(\ln 3)^{k + 1} 3^{2.1}}{(k + 1)!}(0.1)^{k + 1} = \frac{3^{2.1} (0.1\ln 3)^{k + 1}}{(k + 1)!}
$$
How large of $k$ do you need to make $R_k(2.1, 2) < 0.1$?
Calculate the Taylor polynomial of that degree.
$$
\begin{align}
T_k(x) &= a_0 + a_1(x - 2) + \cdots + a_k(x - 2)^k \\
f(2.1) \approx T_k(2.1) &= a_0 + a_1(0.1) + \cdots + a_k(0.1)^k
\end{align}
$$
