Let c be a fixed number.Show that a root of the equation x(x+1)(x+2)...(x+2009)=c can have multiplicity at most 2.Determine the number of values of c for which the equation has a root of multiplicity 2.
I did the first part,I took f(x)=x(x+1)(x+2)...(x+2009)-c then according to the rule I differentiated f(x) and got 2009*x^2008+(1+2+...+2009) which shows it can have a multiplicity of at most 2,as after further differentiation it wont be a polynomial any more. But I cannot do the second part.
A triple (or higher) root of $f(x)=c$ with $f(x)=\prod_{k=0}^{2009}(x+k)$ is also a root of $f'(x)=0$ and $f''(x)=0$. We can exclude the case $c=0$ as then we see immediately, that $f(x)=0$ has $2010$ distinct sinmple roots given by its linear factors. Therefore assume $c\ne0$ from now on. Especially, $f(x)=c$ implies $x\notin\{0,-1,\ldots,-2009\}$. By the product rule and using logarithmic differentiation, $$\tag1\frac{f'(x)}{f(x)} = \sum_{k=0}^{2009}\frac1{x+k}$$ for all $x\notin\{0,-1,\ldots,-2009\}$ and the derivative of this quotient is $$\tag2\frac{f''(x)f(x)-f'(x)^2}{f(x)^2}=-\sum_{k=0}^{2009} \frac1{(x+k)^2}.$$ For $x\notin\{0,-1,\ldots,-2009\}$), the sum on the right is a sum of nozero squares, hence nonzero, i.e. the denominator on the left is nonzero. This implies that $f'(x)=f''(x)=0$ is not possible for such $x$.