Find the error between $f(x) = \sqrt{x}$ and the Taylor Series $T_2(x)$ at $a=100$ on the interval $[90,110]$
I found $T_2(x) = 10 + \frac{1}{20}(x-100)-\frac{1}{500}(x-100)^2$
For the error, I have that:
$|\sqrt{x} - T_2(x) | = \frac{f^{'''}(c)}{3!}|x-10|^{3} $ $\leq$ $\frac{sup_{c \in [90, 110]}f^{'''}(c)}{3!}|x-100|^3$
Since $f^{'''}(x)$ is decreasing, for $sup_{c \in [90, 110]}$, $c = 90$
once I plug this in above, I am not sure how to solve for the error.
I think that your series expansion is not correct. The general expansion is $$\sqrt x=\sum_{n=0}^\infty 10^{1-2 n} \binom{\frac{1}{2}}{n} (x-100)^n$$ and then $$T_3(x)=10+\frac{(x-100)}{20}-\frac{(x-100)^2}{8000}+\frac{(x-100)^3}{1600000}+O\left((x-100) ^4\right)$$ $$\big|\sqrt{x} - T_2(x) \big| = ???$$