Answer
$e^x\approx1+x+\frac{x^2}{2}$
Work Step by Step
The Taylor series for $e^x$ with $r$ in $(-\infty,\infty)$ is
$1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$
We can use the first few terms of a Taylor Series to get an approximate value for a function.
Hence, we can suggest that $e^x\approx1+x+\frac{x^2}{2}$
