Answer
$4.17 \times 10^{-2}$
Work Step by Step
The remainder in the n-th order Taylor's polynomial $T_n(x)$ for a function $f(x)$ centered at $a$ can be expressed as:
$|R_n(x)|=|f(x)-T_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
We are given that $f(x)=e^x$ and $T(x)=1+x+\dfrac{x^2}{2}$ and $n=2$
Thus, we have
$|R_n(x)|\leq 2 \times \dfrac{|\dfrac{1}{2}-0|^{2+1}}{(2+1)!} \approx 4.17 \times 10^{-2}$
