Problem:
Find an upper bound on the error for approximating $f(1/2)$ for $f(x)=\frac{1}{(1+x)^{1/5}}$ using the second degree Maclaurin polynomial.
$|f^3(x)|\leq M$ for $x\in [0,\frac{1}{2}]$
$\frac{66}{125}|\frac{1}{(1+x)^{16/5}}|\leq M$ for $x\in [0,\frac{1}{2}]$
$\frac{66}{125}|\frac{1}{(1+x)^{16/5}}|$ is maximized on $[0,\frac{1}{2}]$ at $x=0$
Therefore, $|R_2(x)|\leq \frac{|f^3(0)|}{3!}|0-0|^3=0$ on $[0,\frac{1}{2}]$ but on desmos, this isn't true.
HINT
By Taylor's theorem, the error is given by $$ E = \sum_{n=3}^\infty \frac{f^{(n)}(1/2)}{n!} (x-1/2)^n = \frac{f^{(n)}(a)}{n!} (x-1/2)^3, $$ where $a \in (0,1/2)$. According to your claim, you can bound $$ f^{(n)}(a) \le \frac{66}{125}. $$ Can you now finish?