Answer
For $r$ in $(-\infty,\infty)$, the Taylor series for $f(x)=2x^3e^{-3x}$ is
$=2x^3-6x^4+9x^5-9x^6+...+\frac{2x^3(-3x)^n}{n!}+...$
Work Step by Step
We are given $f(x)=2x^3e^{-3x}$
for $r$ in $(-\infty,\infty)$
The Taylor series for $e^x$ is
$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...+\frac{1}{n!}x^n+...$
With $c=2x^3$, the Taylor series for $f(x)=2x^3e^{-3x}$ is
$f(x)=2x^3+2x^3.(-3x)+2x^3\frac{(-3x)^2}{2!}+2x^3\frac{(-3x)^3}{3!}+...+2x^3\frac{(-3x)^n}{n!}+...$
$=2x^3-6x^4+9x^5-9x^6+...+\frac{2x^3(-3x)^n}{n!}+...$
