Taylor Polynomials : Error Function

Question

I have answered the first two questions, but was after some help with part c) of the question. I'm not very sure what it was asking, as I have substituted in 2 to the equation in part b to get 1/6*(x-1)^3 but this is not the answer :/

Answer 1

It seems that the author of the exercise missed the point that $f^{(3)}(x)= \frac{2}{x}$ and, hence, it does not contain any $\ln x$.

So, for $1<x<2$ you have $1<z<2$ and, consequently, you get as an estimation for $|\frac{f^{(3)}(z)}{3!}| = \frac{1}{3z}$: $$\frac{1}{3} >\frac{1}{3z} > \frac{1}{6}$$ So, the upper bound in question is $$\frac{1}{3}(x-1)^3$$