Prove there exists a number satisfying an equation.

Question

I have the following problem. Let the function $f(x)=e^x$, $x \in [0,1]$. If $q \in \mathbb{P}_2$ such that $q(0)=f(0)$, $q(1)=f(1)$ and $q'(1)=f'(1)$. Find $q$, prove it is unique and prove that for all $x \in [0,1]$ there exists a $\xi \in (0,1)$ such that $$f(x)-q(x)=\frac{1}{6} x(x-1)^2 e^{\xi}$$. Now finding $q$ and proving its uniqueness is trivial $$q(x)=x^2+(e-2)x+1$$, but it is the last part that confuses me, I noticed that the last expression has $0$ and $1$ as roots and its derivative has $1$ as a root all with respect to $x$, is this useful? How can I prove this?

Answer 1

Looks like you want to adapt the standard proof of the error term in Lagrange’s interpolation polynomial, where the points are $0,1,1$ (as stated e.g. on Wikipedia: https://en.wikipedia.org/wiki/Polynomial_interpolation):

Define $R(x)=f(x)-q(x)$. For a chosen $x\in(0,1)$ define a new function on $[0,1]$ as $Y(t)=R(t)-\frac{R(x)}{x(x-1)^2}\cdot t(t-1)^2$. Differentiate three times: $Y^{(3)}(t)=e^t-6\frac{R(x)}{x(x-1)^2}$. As $Y(0)=Y(x)=Y(1)=Y’(1)=0$, by repeatedly using Rolle’s theorem, we conclude that there is a point $\xi\in(0,1)$ such that $Y^{(3)}(\xi)=0$, which means $e^\xi=6\frac{R(x)}{x(x-1)^2}$ i.e. $R(x)=\frac{1}{6}x(x-1)^2e^{\xi}$ Q.E.D.