Here is an example from my Numerical Analysis book (Burden & Faires). Trying to understand Lagrange error, but I do not understand the statements in bold.
In example 2 we found the second Lagrange polynomial for $f(x)=x^{-1}$ on [2,4] using the nodes $x_0 = 2$, $x1 = 2.75$, $x_2 = 4$. Determine the error form for this polynomial, and the maximum error when the polynomial is used to approximate $f(x)$ for $x\in [2,4]$.
Because $f(x) = x^{-1}$, we have
$$f^{'}(x) = -x^{-2}, f^{''}(x) = 2x^{-3}, and f^{'''}(x) = > -6x^{-4}.$$
As a consequence, the second Lagrange polynomial has the error form
$${f^{'''}(\xi(x))\over 3!}(x - x_0)(x - x_1)(x - x_2) = > -(\xi(x))^{-4}(x-2)(x-2.75)(x-4),\space for\space\xi(x)\in (2,4).$$
The maximum value of $\mathbf{(\xi(x))^{-4}}$ on the interval is $\mathbf{2^{-4}=1/16}$. We now need to determine the maximum value on this interval of the absolute value of the polynomial $$\mathbf{g(x)=(x-2)(x-2.75)(x-4)}$$
Also, what happened to $(\xi(x))^{-4}$ in $g(x)$?
Since $\xi(x) \in (2,4)$, we know that $\xi(x) >2$, which means that $(\xi(x))^{-4} < 2^{-4}$. So, we have an upper bound on the first factor. Next, we're going to worry about getting an upper bound on the second factor, $g(x)$. After we've done that, we will have an upper bound on their product.