Let $f$ be a $C^2[x_0,x_1]$ function and consider the interpolation polynomial of degree $1$ $$p(x_0)=f(x_0)\quad p'(x_1)=f'(x_1)$$ find the error $f(x)-p(x)$ for $x\in[x_0,x_1]$.
I found the polynomial by differentiating $p(x)=ax+b$ and it's $$p(x)=f(x_0)+f'(x_1)(x-x_0)$$ now this looks like the linear term of the taylor series about $x_0$ but that would require $f'(x_0)$. How do I estimate the error for this polynomial? is there a general way?
The usual way to estimate error in this kind of situation is to choose any $x\in(x_0,x_1)$ and then set $$e(y)=f(y)-p(y)-\frac{\left(x_1-y\right)^2-\left(x_1-x_0\right)^2}{\left(x_1-x\right)^2-\left(x_1-x_0\right)^2}\left(f(x)-p(x)\right)$$ Then $e\left(x_0\right)=f\left(x_0\right)-p\left(x_0\right)=0$, $e(x)=f(x)-p(x)-\left(f(x)-p(x)\right)=0$, and $e^{\prime}\left(x_1\right)=f^{\prime}\left(x_1\right)-p^{\prime}\left(x_1\right)=0$, so by Rolles' theorem there is some $c\in\left(x_0,x\right)$ such that $e^{\prime}(c)=0$. Then a further application of Rolles' theorem tells us that there is some $\xi\in\left(c,x_1\right)\subset\left(x_0,x_1\right)$ such that $$e^{\prime\prime}(\xi)=f^{\prime\prime}(\xi)-\frac{2!}{\left(x_1-x\right)^2-\left(x_1-x_0\right)^2}\left(f(x)-p(x)\right)=0$$ Thus for any $x\in\left(x_0,x_1\right)$ there is some $\xi\in\left(x_0,x_1\right)$ such that $$f(x)=p(x)+\frac{\left(x_1-x\right)^2-\left(x_1-x_0\right)^2}{2}f^{\prime\prime}(\xi)$$ The key step was finding a nontrivial polynomial of degree one higher than the interpolating polynomial that matched all the boundary conditions for the function $f(x)=0$. Once you've got that the rest follows mechanically.
EDIT Since the first derivative was also matched there might be some question about the error in the first derivative. To resolve this issue we take the first derivatives of the numerator and denominator functions of $x$ to get $$g(y)=f(y)-p(y)+\frac{\left(x_1-y\right)^2-\left(x_1-x_0\right)^2}{2\left(x_1-x\right)}\left(f^{\prime}(x)-p^{\prime}(x)\right)$$ So that $g\left(x_0\right)=g^{\prime}(x)=g^{\prime}\left(x_1\right)=0$ and it follows directly that $$f^{\prime}(x)=p^{\prime}(x)-\left(x_1-x\right)f^{\prime\prime}(\xi)$$ For any $x\in\left(x_0,x_1\right)$ for some $\xi\in\left(x,x_1\right)$.