Using a Taylor polynomial of degree 2, find an approximation for $\sqrt[3]{e}$ I do not understand how to find an approximation using a Taylor polynomial.
Also, I need to find an upper limit to the remainder $|R_2(x)|$.
Excuse me if I have any obvious mistakes, this is my first time solving something like this.
$$R_2(\frac{1}{3}) = \frac{f^{(3)}(c)}{3!}(\frac{1}{3})^3$$
$0 \le c \le x$ and $f^{(3)}(c) = e^c$, therefore:
$$= \frac{e^c}{6}\cdot\frac{1}{27} \le \frac{e^\frac{1}{3}}{162}$$
I have no idea. This is probably completely wrong, please help.

You have
$$ e^{x} = \sum_{n\geq 0}\frac{x^n}{n!} $$
for any $x\in\mathbb{C}$, hence $\sqrt[3]{e} = \sum_{n\geq 0}\frac{1}{3^n n!}$, which is a fast-convergent series. You may notice that the terms associated to $n\geq 4$ are $\ll\frac{1}{1000}$, hence it looks reasonable that the partial sum up to $n=3$ is already an approximation of $\sqrt[3]{e}$ within the required accuracy, i.e.
$$ \sqrt[3]{e} \approx \sum_{n=0}^{3}\frac{1}{3^n n!}=\frac{113}{81}=\color{green}{1.395}06\ldots $$
The error of such estimation can be evaluated through standard results on the magnitude of $f(x)-P(x)$, where $P(x)$ is a Taylor polynomial for $f$ (centered at the origin), or just "by hand"
$$ \sum_{n\geq 4}\frac{1}{3^n n!} \leq \sum_{n\geq 4}\frac{1}{3^n\cdot 5^{n-4}\cdot  4! }=\frac{5}{9072}< 6\cdot 10^{-4}$$
through geometric series. An interesting alternative approach is to notice that
$$\frac{1}{4^4}\geq \int_{0}^{1} x^4(1-x^4) e^{x/3}\,dx = 5832\left(114961\sqrt[3]{e}-160441\right)$$
hence
$$ \sqrt[3]{e} \approx \frac{160441}{114961} $$
with an error less than $10^{-10}$.
