## University Calculus: Early Transcendentals (3rd Edition)

a) $Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$ b) $0 \lt x \lt 0.21544$
a) The Taylor series for $f(x)$ at $x=0$ is as follows: $Q(x)=f(0)+f^{,}(x-0)+\dfrac{f^{,}(0)}{2!}(x-0)^2=1+kx+\dfrac{k(k-1)}{2}x^2$ b) Consider the Remainder Estimation Theorem to find the Error. So, $|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$ and $|R_2(x)| =|\dfrac{ 3! x^3}{3 !}|=|x^3|$ Since, $|x| \lt \dfrac{1}{100}$, thus, $|x|^3 \lt \dfrac{1}{100}$ or, $\dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10}$ or, $0 \lt x \lt \dfrac{1}{(100)^{1/3}}$ or, $0 \lt x \lt 0.21544$ Thus, the required values of $x$ are $0 \lt x \lt 0.21544$ . a) $Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$ b) $0 \lt x \lt 0.21544$