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