# If $f(x)$ has $n$ distinct roots in $R$, then $f'(x)$ has $n-1$ distinct roots in $R$ Without Rolle's Theorem

Prove that: If $$f(x)$$ has $$n$$ distinct roots in $$R$$, then $$f'(x)$$ has $$n-1$$ distinct roots in $$R$$ Without Rolle's Theorem.

I know in this topic, There is proof with Rolle's theorem. It uses that if we sort the roots such that $$x_i < x_{i+1}$$,then $$\exists c_i\in(x_i,x_{i+1}) ; f'(c_i) = \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i} =0$$ where $$i = \{ 1 , 2 ,... , n-1\}$$.

But I want to know is there any way to prove this without Rolle's theorem. Maybe using Intermediate value theorem. if there isn't repetitive roots, I can prove that with intermediate value theorem. (i.e. the power of each term in factorization of $$f(x)$$ is $$1$$). In this solution, I find that $$sgn( f'(x_1)) = (-1)^{n-1},~~ sgn(f'(x_2)) = (-1)^{n-2} , ...$$ ($$x_i$$ are sorted) and I find distinct roots using IVT. But If the power of each root get higher than one, this method doesn't work. So Is there any way to prove that generally without Rolle's Theorem?

By IVT, I mean this: (link to wikipedia)

• @qbert: Yes,. Sorry for messing up in my comment – I was looking at several problems at the same time. I've deleted this irrelevanty comment. – Bernard Nov 25 '18 at 20:38
• @Bernard no problem, I knew you knew, I just wanted to avoid confusing the OP. – qbert Nov 25 '18 at 20:40
• What is $f$? A polynomial of degree $n$? – Paul Frost Nov 25 '18 at 21:55

Let $$a, b$$ with $$a < b$$ be two consecutive roots of a polynomial $$p(x)$$. Then we have $$p(x) =(x-a) ^m(x-b) ^nq(x)$$ where $$m, n$$ are positive integers and $$q(x)$$ is a polynomial having no roots in $$[a, b]$$. Then $$p'(x) =q'(x) (x-a) ^m(x-b) ^n+\{m(x-b)+n(x-a)\}(x-a)^{m-1}(x-b)^{n-1}q(x)$$ This can be further written as $$p'(x) =(x-a) ^{m-1}(x-b)^{n-1}r(x)$$ where $$r(x) =(x-a) (x-b) q'(x) +\{m(x-b) +n(x-a) \}q(x)$$ Then $$r(x)$$ is a polynomial such that $$r(a) r(b) =-mn(a-b) ^2q(a)q(b)<0$$ (note that $$q$$ maintains a constant sign in $$[a, b]$$ as it has no roots in $$[a, b]$$). It follows from IVT that $$r(x)$$ has a root $$c\in(a, b)$$ and since $$r(x)$$ is a factor of $$p'(x)$$ it follows that $$p'(x)$$ has a root $$c\in(a, b)$$.