# Rolle's Theorem in reverse

I'm trying to figure out the following problem:

If $\sum_{i=0}^n\frac{a_i}{i+1}=0$, then prove $\sum_{i=0}^n a_i x^i = 0$ has a solution in $(0,1)$.

I'm trying to do the following:

Let $f(x) = \sum_{i=0}^n \frac{a_i x^{i+i}}{i+1} \implies f'(x) = \sum_{i=0}^na_ix^i, f(1) = \sum_{i=0}^n\frac{a_i}{i+1}$. And I want to somehow use Rolle's theorem in reverse. My understanding is that to use Rolle's theorem normally, I would show that $f$ is continuous on $[0,1]$ and differentiable on $(0,1)$ where $f(a) = f(b)$ which then implies there is a solution in $(0,1)$. However, I am starting with assuming there is a solution in $f'$ and trying to show there is a solution in $f$. I feel like I need to somehow use Rolle's in reverse, but I'm not sure how to show this.

• You're not starting by assuming that there is a solution for $f'(x) = 0$, that would be your conclusion. Don't worry, you wouldn't be using Rolle's theorem backwards. The only thing that is "backwards" is that you had to write down a function $f(x)$ that you wanted to apply Rolle's theorem to, even though $f(x)$ wasn't given to you at the outset, and you're drawing a conclusion about $f'(x)$, which was given to you. Are there any of the hypotheses of Rolle's theorem that are not clearly satisfied here? – user49640 May 3 '17 at 4:51
• $f$ as you have defined it is a polynomial, so it is continuous and differentable. $f(0) = 0$ and $f(1) = 0$. So what does Rolle's theorem say about $f'(x)$ in $(0,1)$? – Doug M May 3 '17 at 5:11

## 1 Answer

You are nearly there: $f(0)=0=f(1)$. So just apply Rolle to $f$.