I want to find the number of possible $c \in \mathbb R$ such that the polynomial $$ f(x) = x(x+1)(x+2)\dots(x+2011) - c $$ has a double root ( root of multiplicity 2 ).
I know that this corresponds to finding the number of distinct values of extrema of the function $$ g(x)= x(x+1)(x+2)\dots(x+2011) $$ after which I must set $c$ to be equal to the value of the extrema.
I thought of using the symmetry about the point $x=-1005.5$ to show that (apart from the central extremum), distinct values of extrema occur in pairs.
What remains to prove is that apart from symmetric pairs, all critical points have distinct values of $f(x)$
I have looked at the polynomials $x(x+1)(x+2)(x+3)$ and $x(x+1)(x+2)(x+3)(x+4)(x+5)$ and they seem to agree with my hypothesis but I am unable to prove it for the general case $$ P(x) = x(x+1)(x+2)(x+3)\dots (x+2k+1) $$
The roots of $g$ are at $0,-1,-2,\ldots,-2011$. There must be a critical point between each adjacent pair of roots, so each of the intervals $(-1,0)$, $(-2,-1)$, $\ldots$, $(-2011,-2010)$ contains a critical point. This accounts for all the critical points.
Observe that $g(x)$ satisfies the functional equation $$ g(x-1)=\frac{x-1}{x+2011}g(x). $$ The rational function factor on the right hand side has absolute value smaller than $1$ if $x>-1005$, so for these $x$ we have $|g(x-1)|<|g(x)|$. This means the critical values in the intervals $(-1,0)$, $\ldots$, $(-1006,-1005)$ are strictly decreasing in absolute value. So we have found $1006$ distinct critical values.
As you have observed, $g$ satisfies the symmetry property $g(x) = g(-2011-x)$. This means the the critical values come in pairs, so the $1006$ critical values above are all of the distinct values.