How many solutions to $f'(x)=0$

Question

How many solutions to $f'(x)=0$, when $f(x)=(x-1)(x-2)...(x-n)$

I know that $f$ is a polynomial of degree $n$, so $f'$ has at most $n-1$ roots It depends on whether $n$ is odd or even ?

Thanks

Answer 1

Since $f(x)=0$ for $x\in\{1,\ldots,n\}$ we have by Rolle's Theorem $f'(x_k)=0$ for suitable $x_k\in(k-1,k),k\in\{2,\ldots,n\}$, that is, $f'(x)=0$ has hat least $n-1$ solutions. But, as you already said, there are at most $n-1$ solutions, since $f'$ is a polynomial of degree $n-1$, we have exactly $n-1$ solutions.

Answer 2

It will always be $n - 1$ for the given polynomial type. Try plotting the graph of the function for both cases, $n$ even or odd.