MHB Intermediate Value Thm for Five-Point Formula

[kalish1]

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

http://www.math.ucla.edu/~yanovsky/Teaching/Math151A/hw5/Hw5_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.

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

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

Shouldn't it be

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

 

[Chris L T521]

Hi kalish,

Here's a friendly reminder about this part of Rule #2 here on MHB.

As a courtesy, if you post your problem on multiple websites, and you get a satisfactory response on a different website, indicate in your MHB thread that you got an answer elsewhere so that our helpers do not duplicate others' efforts.

As a courtesy (not just to us, but also to those on other sites as well), you should inform our (and their) members when you've posted a question on multiple sites (for instance, I see that you've asked this same question on math.stackexchange) and then inform us if you find a solution elsewhere. This way, no one's efforts are duplicated and/or put to waste on solving a problem that has been potentially solved. (Smile)

 

[kalish1]

Hi kalish,

Here's a friendly reminder about this part of Rule #2 here on MHB.
As a courtesy (not just to us, but also to those on other sites as well), you should inform our (and their) members when you've posted a question on multiple sites (for instance, I see that you've asked this same question on math.stackexchange) and then inform us if you find a solution elsewhere. This way, no one's efforts are duplicated and/or put to waste on solving a problem that has been potentially solved. (Smile)

Thanks Chris. I decided that this problem was unnecessary because the goal I am trying to achieve doesn't actually make sense. So I would like to close this post.