Answer
$2.60 \times 10^{-4}$
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:
$|R_n(x)|=|f(x)-T_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}~~~(1)$
We are given the functions $f(x)=e^{-x}$
and $|f''(c)|=|e^{-x}| \leq e^0 =1 \implies M=1$
Put these values in the equation (1) to obtain:
$|R_{n}(x)| \leq 1 \dfrac{|x-a|^{n+1}}{(n+1)!}$
Thus, we have: $|R_{4}(0.5)| \leq 1 \dfrac{|0.5-0|^{4+1}}{(4+1)!}=2.60 \times 10^{-4}$
