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
$$f(0)=0,; f(1)=M,;ftext{ is continuous on }[0,1],; ftext{ is differentiable on }(0,1),;text{ and }xitext{ is the unique point strictly between 0 and 1 such that};f'(xi)=M,.$$

For the $Mneq 0$ and $xineq 1/e$ 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 $Mneq 0$ and $xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+xln(x))$.

For $M=0$, we first choose $alphageq 1$ and $betageq 1$ 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).

First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.

Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.

If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.

For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.

Explanation

Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.

$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.

And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.

This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)

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 $xiin (0,1),$ 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 $Mne 0$ 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 $Mne 0.$

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 $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$

For $0le b le 1,$ 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 $bin [0,1)$ 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 $xiin [1/2,1/2^{1/2}],$ 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 $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$

What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$

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 $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ 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 $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$

We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ 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}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$