$$\frac{(x_1−x_0)^2}{8}M_2 : \{M_2= \max\{|f''(x)|, x \in[x_0,x_1]\}$$
I know that that part of the formula that goes $(x-x_0)(x-x_1)...$ is a 2nd degree polynomial, and I know that the maximum point is given by $\frac{x_1-x_0}{2}$, no troubles there. I'm just having trouble working out the rest. I did:
$$(\frac{x_1-x_0}{2}-x_0)(\frac{x_1-x_0}{2}-x_1) =\\ \frac{1}{4}(x_1-3x_0)(-x_1-x_0) = \\\frac{1}{4}(-x_1^2-x_0x_1+3x_0x_1+3x_0^2) = \\ -\frac{1}{4}(x_1^2-2x_1x_2-3x_0^2)$$
That $-3x_0^2$ isn't right. It should be $+x_0^2$. I don't see where I made a mistake. Help?
For linear interpolation
$$f(x) - P_1(x) = \dfrac{(x-x_0)(x-x_1)}{2} f''(c_x)$$
The error bound is
$$E_1 = \tag 1 |f(x) - P_1(x)| \le \max_{x_0 \le x \le x_1} \left| \dfrac{(x-x_0)(x-x_1)}{2}\right| \max_{x_0 \le x \le x_1}|f''(x)|$$
Finding $\displaystyle \max_{x_0 \le x \le x_1} \frac{(x-x_0)(x-x_1)}{2}$
$$\dfrac{\partial}{\partial{x}} \left[(x-x_0) (x- x_1)\right] = 2x - x_0-x_1 = 0 \implies x = \dfrac{x_0+x_1}{2}$$
Substituting this max $x$
$$\left| \dfrac{(x-x_0)(x-x_1)}{2}\right| = \left|\left(\dfrac{x_0+x_1}{2} - x_0\right)\left(\dfrac{x_0+x_1}{2} - x_1\right)\right| = \left|-\frac{1}{4} (x_1 - x_0)^2\right|$$
Note
$$\left(\dfrac{x_0+x_1}{2} - x_0\right)\left(\dfrac{x_0+x_1}{2} - x_1\right) = -\frac{x_0^2}{4}+\frac{x_0x_1}{2}-\frac{x_1^2}{4} = -\frac{1}{4} (x_1-x_0)^2$$
Substituting back into $(1)$
$$E_1 = |f(x) - P_1(x)| \le \dfrac{(x_1-x_0)^2}{8} \max_{x_0 \le x \le x_1}|f''(x)|$$