I'm wondering if there's in a mistake in either my reasoning or the given solution for the problem and was hoping to have someone double check this for me.
The problem states:
Let $g(2)=3$ , $g'(2)=1$ , $g''(2)=−1$, and $g'''(2)=2$ . Use the second-degree Taylor polynomial for $g'$ centered at $x=2$ to approximate $g'(1)$ .
The answer states:
Taylor polynomial of degree 3 for $g$ centered at $x=2$ is given by:
$3+1(x-2)-\frac{1(x−2)^2}{2!}+\frac{2(x−2)^3}{3!}$
Hence, the second-degree Taylor polynomial for $g'$ can be found by taking the derivative of the above expression. $1−(x−2)+\frac{(x−2)^2}{2}$
Therefore, $g ′(1)≈1+1+\frac{1}{2}=\frac{5}{2}$
However, I calculated the derivative as
$1−(x−2)+(x−2)^2$, which gives me $g'(1) = 3$. When taking the derivative of the 3rd degree Taylor polynomial, it seems like the coefficient on the last term should cancel with the $3!$ in the denominator, but I might have overlooked something.