Polynomial Interpolation My professor gave the following question as a practice for study guide. Any assistance in terms of helping me to solve this would be much appreciated. 
Suppose that $f$ is continuous and has continuous first and second derivative on the interval $[x_0,x_1]$. Derive the following bound on the error due to linear interpolation of $f$
$$|f(x)-P_1(x)|<=1/8 h^2 max_{x \in[x_0,x_1]}|f''(x)|$$
where $h=x_1-x_0$.
 A: For the upper bound on the second derivative, draw a graph of a linear interpolation on an interval $[x_0,x_1]$ and draw a function that has numerous local maximum and minimums on the interval. 
Now see if you can derive a relationship between the maximum error between the interpolation and the function and the curvature of the second derivative, using a two term taylor series expansion of the function where the remainder is your upper bound for the second derivative. After looking at the geometry of the problem you should be able to derive the following relation
$$
\left|f(x)-P_1(x)\right|\le(x-x_0)(x-x_1)\frac{M}{2!}
$$
where the last term comes from the Taylor's series expansion, where M is the upper bound of the second derivative. For the first term $(x-x_0)(x-x_1)$ think about how it is related to h through the geometry of the problem and what the upper bound of h must satisfy. If you get stuck with the last bit have a look here (page 4)
Goodluck!
A: For $f(x)$ you get by the fundamental theorem and partial integration
\begin{align}
f(x)&=f(x_0)-(x-x_0)f'(x)+\int_{x_0}^x (s-x_0) f''(s)\,ds\\
f(x)&=f(x_1)+(x_1-x)f'(x)+\int_x^{x_1} (x_1-s) f''(s)\,ds
\end{align}
and by distributivity
$$h\,f(x)=(x_1-x)f(x)+(x-x_0)f(x)$$
by careful combination of these equations, the first derivatives cancel, the constant terms combine to give the linear interpolation and the integrals can be estimated to give the error term.
\begin{align}
f(x)&=P_1(x)+\int_{x_0}^{x_1}\frac{(x_1-\max(x,s))(\min(x,s)-x_0)}{x_1-x_0}f''(s)\,ds
\\
&=P_1(x)+f''(\xi_x)\frac{(x_1-x)(x-x_0)}2~~~\text{ for some }\xi_x\in[x_0,x_1],
\end{align}
using the integral mean value theorem. Now apply the mean inequality $(x_1-x)(x-x_0)\le\frac{(x_1-x_0)^2}4$
