Use the zeroes of $\bar{T}_3$ and transformations of the given interval to construct an interpolating polynomial of degree 2 for $f(x)={ 1\over x}$ over the interval $[1,3]$
My biggest issue is finding the zeroes. How exactly do I do that? I have been using the formula $\frac{1}{2}\left[(b-a)\cos\left(\frac{N+\frac{1}{2}-n}{ N}\cdot\pi\right) +a+b\right]$ Where $[a,b]$ is the interval, $N$ is the degree and for $n=1,2,3,\ldots, N.$ $\bar{x_k}=\frac{1}{2}\left[(3-1)\cos\left(\frac{2+\frac{1}{2}-1}{2}\cdot\pi\right) +1+3\right]=1.29289322$ This is of course wrong. If anyone is familiar with this topic then please guide me.
The final answer is $P(x)=.3489153-.1744576(x-2.866025)+.1538462(x-2.866025)(x-2)$.