<s>[INST] Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem. I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:

For each real number $M$ and each real number $\xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
  $$f(0)=0,\; f(1)=M,\;f\text{ is continuous on }[0,1],\; f\text{ is differentiable on }(0,1),\;\text{ and }\xi\text{ is the unique point strictly between 0 and 1 such that}\;f'(\xi)=M\,.$$

For the $M\neq 0$ and $\xi\neq 1/e$ case, we can show that
$$g(x)=\begin{cases}
0&\text{ if }x=0,\\
1/e&\text{ if }x=1\\
1&\text{ if }x=\infty,\\
\sqrt[1-x]{x}&\text{ otherwise}
\end{cases}$$
is strictly increasing and continuous on $[0,\infty]$. Thus there is a unique positive $\alpha$ such that $g(\alpha)=\xi$. In turn, we can define $f(x)=Mx^\alpha$ which will satisfy the claim. For the $M\neq 0$ and $\xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+x\ln(x))$.
For $M=0$, we first choose $\alpha\geq 1$ and $\beta\geq 1$ such that $\frac{\alpha}{\alpha+\beta}=\xi$. We then define $f(x)=x^\alpha(1-x)^\beta$ which will satisfy the claim.
My question however is this:

Can we construct such an $f$ to be a polynomial?

An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $\xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
 [/INST] Robert's comment about cubics is the best we can do.

Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $\xi \in (0, 1)$ for which $f'(\xi) = 0$, then $\frac{1}{3} \leq \xi \leq \frac{2}{3}$.

Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $\xi = \frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $\xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(\xi) = 3\xi^2 + 2k \xi - (1+k) = 0$ are thus $$\xi = \frac{k \pm \sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 \xi - k)^2 = k^2 + 3k + 3$, or $$k = \frac{1 - 3 \xi^2}{2 \xi - 1}$$ but the squaring means that $\xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $\xi$, up to scaling of the coefficients, is $$f(x) = (2 \xi - 1) x^3 + (1 - 3 \xi^2) x^2 + (3 \xi^2 - 2\xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 \xi - 1) x^2 + 2(1 - 3 \xi^2) x + (3 \xi^2 - 2\xi) = 0$ add up to $\frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)}.$ If $\xi$ is one solution, then the other solution (call it $\xi'$) is \begin{align*} \xi' &= \frac{2 (3 \xi^2 - 1)}{3 (2 \xi - 1)} - \xi \\ &= \frac{6 \xi^2 - 2}{6 \xi - 3} - \frac{6 \xi^2 - 3 \xi}{6 \xi - 3} \\ &= \frac{3 \xi - 2}{6 \xi -3} \\
&= \frac{3 \xi - \frac{3}{2}}{6 \xi - 3} - \frac{\frac{1}{2}}{6 \xi - 3} \\
&= \frac{1}{2} - \frac{1}{12\xi - 6}.\end{align*}
Thus, $\xi' \notin (0, 1)$ if and only if $|12 \xi - 6| \leq 2$, i.e., if $\frac{1}{3} \leq\xi \leq \frac{2}{3}.$
</s>