Proof that $ f'(x)=0 $ has $n-1$ roots

Question

I am given $f(x)=(x-1)(x-2)(x-3)...(x-n)$. I am obliged to proof that for $n \ge 1$, the equation $ f'(x)=0 $ has $n-1$ roots.

I think I have to somehow use Rolle's theorem to proof that. I will be glad for any help and tips.

score 8 · Accepted Answer · answered Jan 27 '14 at 23:30

8

Hints:

The original function has exactly $n$ roots, at $1, 2, 3, ..., n$.
Consider Rolle's Theorem on the intervals $[1, 2]$, $[2, 3]$, ..., $[n - 1, n]$.
The derivative is a polynomial of degree at most $n - 1$.

Can you see how to put it together?

answered Jan 27 '14 at 23:30

When I consider Rolle's Theorem on the intervals I am able to say that there are exactly $n-1$ points $c$ that $f'(c) = 0 $ But I do not understand that it implies that $f'(x)$ on $\mathbb{R}$ has $n-1$ roots?. – Marcin Majewski Jan 27 '14 at 23:34
Nevermind, I think I understand this. Thank you – Marcin Majewski Jan 27 '14 at 23:37
@MarcinMajewski You're very welcome. – Jan 27 '14 at 23:37

ziang chen · Answer 2 · 2014-01-28T11:04:57.870

4

$f(k)=f(k+1)$, $k=1,2,\dotsc,n-1$, so there exists a $\alpha_k\in(k.k+1)$ such that

$$f'(\alpha_k)=0$$

edited Jan 28 '14 at 11:04

answered Jan 27 '14 at 23:32

ziang chen

2 Answers2