Answer
$1.6 \times 10^{-5}$
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 $|f^{n+1}(x)|=|e^x| \leq e^{0.25} \leq 2$
Put these values in the equation (1) to obtain:
$|R_{n}(x)| \leq 2\dfrac{|x-a|^{n+1}}{(n+1)!}$
Thus, we have: $|R_{n}(0.25)| \leq 2 \dfrac{|0.25-0|^{4+1}}{(4+1)!}=1.6 \times 10^{-5}$
