Solution
= Solution
The <Taylor polynomial> uses $f(1)=1/2$, $f'(1)=-1/2$ and $f''(1)=1/2$. Therefore
$$
\boxed{f_2(x)=\frac12-\frac12(x-1)+\frac14(x-1)^2=\frac{x^2-4x+5}{4}.}
$$
The quadratic coefficient is $f''(1)/2!$, not $f''(1)$.
= Solution
The <Taylor polynomial> uses $f(1)=1/2$, $f'(1)=-1/2$ and $f''(1)=1/2$. Therefore
$$
\boxed{f_2(x)=\frac12-\frac12(x-1)+\frac14(x-1)^2=\frac{x^2-4x+5}{4}.}
$$
The quadratic coefficient is $f''(1)/2!$, not $f''(1)$.