Answer
$6.4 \times 10^{-7}$
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 $f(x)=\ln (1+x)$
and $|f^4(c)|=|\dfrac{6}{(1+c)^4}| \leq 6 \implies M=6$
Put these values in the equation (1) to obtain:
$|R_{n}(x)| \leq 6 \dfrac{|x-a|^{n+1}}{(n+1)!}$
Thus, we have: $|R_{3}(0.04)| \leq 6 \dfrac{|0.04-0|^{3+1}}{(3+1)!}=6.4 \times 10^{-7}$
