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I'm reading these set of online notes and it reads as following:

  1. $f(x)- P(x)$=$\frac{(x-x_0)(x-x_1)(x-x_2)}{3!}$$f'''c(x)$ where $c(x)$ is some point between the minimum and maximum of the points in ${x, x_0, x_1, x_2}$.
  2. Then they say: Let $x_1=x_0+h$, $x_2=x_1+h$.
  3. Denote $\phi_2(x)=(x-x_0)(x-x_1)(x-x_2)$
  4. We must compute:If we want a uniform bound for $x_0 ≤ x ≤ x_2$, we must compute

$\max\limits_{x_0\leq x\leq x_2}$ $|\phi_2(x)|$ = $\max\limits_{x_0\leq x\leq x_2}$ $|(x − x_0) (x − x_1) (x − x_2)| $.

  1. Using Calculus:

$\max\limits_{x_0\leq x\leq x_2}$ $|\phi_2(x)|$=$\frac{(2h^3)}{3\sqrt3}$ at $x=x_1 \pm$$\frac{(h)}{\sqrt3}$ .

I'm confused by Step 5. How did they determine that the maximum was at $x=x_1 \pm$$\frac{(h)}{\sqrt3}$ . I tried setting the derivative equal to $0$ but I can't seem to get the same result. The confusing thing is that you can write $x_0$,$x_1$, $x_2$ in terms of h. So I'm not sure if they converted $\phi_2(x)$ in terms of $x_1$'s and $h$'s. Any help would be much appreciated. Sorry for the basic problem, I just really need to know how they derived step 5. Thank you very much.

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Take the derivative of the expression $(x-x_0)(x-x_1)(x-x_2)$, and equate it to $0$. $$(x-x_0)(x-x_1)+(x-x_0)(x-x_2)+(x-x_1)(x-x_2)=0$$ We know that $x_0=x_1-h$ and $x_2=x_1+h$, so we have $$\begin{align}[(x-x_1)+h](x-x_1)+[(x-x_1)+h][(x-x_1)-h]+(x-x_1)[(x-x_1)-h]=0\\(x-x_1)^2+h(x-x_1)+(x-x_1)^2-h^2+(x-x_1)^2-h(x-x_1)=0\\3(x-x_1)^2-h^2=0\\(x-x_1)^2=\frac{h^2}{3}\\x=x_1\pm\frac{h}{\sqrt 3}\end{align}$$

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  • $\begingroup$ I think the negatives are a bit off the first expression should be x-x1+h not (x-x1)-h $\endgroup$ – rain Jun 27 '18 at 1:15
  • $\begingroup$ Correct. Just happens that I have both positive and negative terms, so since I've missed all terms, I get the correct answer. Will fix it anyway. Thanks $\endgroup$ – Andrei Jun 27 '18 at 1:20

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