Answer
$e^{-x}\geq 1-x$ for all x
Work Step by Step
The Taylor series for $e^{-x}$ with $r$ in $(-\infty,\infty)$ is
$1+(-x)+\frac{1}{2!}(-x)^2+\frac{1}{3!}(-x)^3+...+\frac{1}{n!}(-x)^n+...$
$=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...+\frac{1}{n!}(-x)^n+...$
$e^{-x}=(1-x)+(\frac{x^2}{2!}-\frac{x^3}{3!})+...+\frac{1}{n!}x^n+...$
$e^{-x}-(1-x)\geq0$
(0 represents $(\frac{x^2}{2!}-\frac{x^3}{3!}+...+\frac{1}{n!}(-x)^n+...)$
$e^{-x}\geq 1-x$
Hence, we can suggest that $e^{-x}\geq 1-x$ for all x
