I'm trying to do test corrections for Calculus 2, but I am stuck on this problem.
Find an upper bound for your error estimate in part (b)
The estimate is of $\sqrt2$ using the 4th degree Taylor series of $(1+x)^{1/3}$ (centered at zero). The esimate was that $\sqrt[3]{2} \approx \frac{302}{243}$ $$T_4(x)=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3-\frac{10}{243}x^4$$
Using the lagrange error $$|E_n(x)|\le\frac{M}{(n+1)!}x^{n+1}$$ Where $M$ is the maximum value of the $n+1^{th}$ derivative of $f(x)$ at the value.
The 5th derivative of $(1+x)^{1/3}$ is $\frac{880}{243}(1+x)^{-14/3}$
Therefore, $$\star \quad M=\frac{880}{243}(2)^{-14/3}\approx0.14258$$ $$M\lt1/2$$ $$|E_4(x)|\le\frac{\frac{1}{2}}{(5)!}1^{5}$$ $$|E_4(x)|\le\frac{1}{240}$$ This is the answer I came up with, but I was marked down two points.
My instructor told me I should have evaluated the 5th derivative at $x=0$ instead of $x=1$ (the starred step). Is he right? Could somebody explain why?