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).
 A: Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $\xi=1/2.$ For other values of $\xi\in (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $\xi$ is given, we take $p=\ln(1/2)/\ln \xi,$ and $g_p$ solves the problem. Finally, if $M\ne 0$ and $\xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $M\ne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $f\circ p$ is a polynomial that solves the problem for for the unique value $\xi \in(0,1)$ such that $p(\xi)=1/2.$
For $0\le b \le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then  $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $b\in [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= \frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $\xi\in [1/2,1/2^{1/2}],$ there is a unique $b_{\xi}\in [0,1]$ such that $p_{b_{\xi}}(\xi)=1/2.$ Verify that $b_{\xi}$ is given by the formula
$$b_{\xi} = \frac{1/2-\xi^2}{\xi(1-\xi)}.$$
So we've solved the problem for $\xi\in[1/2,1/2^{1/2}].$ But we've also solved it for $\xi\in [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $\xi,$ $f\circ p_{b_{\xi^2}}(x^2)$  does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $\xi\in [1/2,1).$
What about $\xi\in (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $\xi\in [1/2,1),$ then $1-g(1-x)$ is a solution for $1-\xi.$ 
A: 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 $\frac{1}{3} \leq\xi \leq \frac{2}{3}.$
