Answer
$f(x)=1+x+x^2+x^3$ and $Error \leq 0.00017$
Work Step by Step
Write the Taylor series for $f(x)$ at $x=0$ as follows: $f(x)=1+x+x^2+x^3$
Apply the Remainder Estimation Theorem to compute $|f^{4} | \leq M$.
$|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
and $|R_3(x)| =|f^{4} |\times |\dfrac{x^4}{4!}| =|\dfrac{x^4}{(1-x)^5}|$
Since, $|x| \lt 0.1$ , we will find the maximum of $|R_3(x)| $, so
$\space max |\dfrac{x^4}{(1-x)^5}|$
So, $|x| \lt 0.1$
and $\dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10} $
Now $Max |\dfrac{x^4}{(1-x)^5}| \lt (0.1)^4 \times (\dfrac{10}{9})^5$
After simplifications, we get:
$Max |\dfrac{x^4}{(1-x)^5}| \lt 0.00016935 \leq 0.00017$
