Answer
The Taylor series for $f(x)=e^{-5x}$
for $r$ in $(-\infty,\infty)$ is
$=1-5x+\frac{25x^2}{2}-\frac{125x^3}{6}+...+\frac{(-5x)^n}{n!}...$
Work Step by Step
We are given $f(x)=e^{-5x}$
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 $f(x)=e^{-5x}$ is
$f(x)=1+(-5x)+\frac{(-5x)^2}{2!}+\frac{(-5x)^3}{3!}+...+\frac{(-5x)^n}{n!}+...$
$=1-5x+\frac{25x^2}{2}-\frac{125x^3}{6}+...+\frac{(-5x)^n}{n!}...$
