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.

Use the Taylor series for $e^x$ evaluated at $x=\frac{1}{3}$:
$$e^{1/3}=\sum_{n=0}^\infty\frac{1}{3^nn!}=1+\frac{1}{6}+\frac{1}{54}+\cdots$$
which is from the general form
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
