Application of Intermediate Value Theorem for five-point formula (numerical different

I have a specific, for-learning-sake-only question on how the author of this link:

http://www.math.ucla.edu/~yanovsky/T..._solutions.pdf

gets past the details of the Intermediate Value Theorem on the following paragraph. If someone could fill in the details for me, it would be greatly appreciated because I'm having a hard time understanding.

\displaystyle \begin{align} \left(\frac{3}{12h}f^{(5)}(\xi_1)+\frac{18}{12h}f^ {(5)}(\xi_2)-32\frac{6}{12h}f^{(5)}(\xi_3)+243\frac{1}{12h}f^{( 5)}(\xi_4)\right)\frac{h^5}{120}&= \\ \left(\frac{3}{12}f^{(5)}(\xi_1)+\frac{18}{12}f^{( 5)}(\xi_2)-32\frac{6}{12}f^{(5)}(\xi_3)+243\frac{1}{12}f^{(5) }(\xi_4)\right)\frac{h^4}{120}&= \\ 6f^{(5)}(\xi)\frac{h^4}{120}&= \\ \frac{h^4}{20}f^{(5)}(\xi) \end{align}

"Note that the IVT was used above..."

Shouldn't it be

"Suppose $\displaystyle f^{(5)}$ is continuous on $\displaystyle [x_0-h,x_0+3h]$ with
$\displaystyle x_0-h<\xi_1<x_0<\xi_2<x_0+h<\xi_3<x_0+2h<\xi_4<x_0+3h$. Since $\displaystyle \frac{1}{4}[f^{(5)}(\xi_1)+f^{(5)}(\xi_2)+f^{(5)}(\xi_3)+f^{(5 )}(\xi_4)$] is between [TEX]f^{(5)}(\xi_1)[\TEX] and $\displaystyle f^{(5)}(\xi_4)$, the Intermediate Value Theorem implies that a number $\displaystyle \xi$ exists between $\displaystyle \xi_1$ and $\displaystyle \xi_4$, and hence in $\displaystyle (x_0-h,x_0+3h)$, with $\displaystyle f^{(5)}(\xi)=\frac{1}{4}[f^{(5)}(\xi_1)+f^{(5)}(\xi_2)+f^{(5)}(\xi_3)+f^{(5 )}(\xi_4)]$"?