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If $$C_{0} + \frac{C_1}{2} + \cdots+ \frac{C_{n-1}}{n} + \frac{C_{n}}{n+1} = 0\,,$$ prove that $$C_{0} + C_{1}x + \cdots+ C_{n-1}x^{n-1} + C_{n}x^{n} = 0$$ has at least one real root between $0$ and $1$.

I know how to prove the result by using the mean value theorem, but I am not understanding how the result from the mean value theorem allows us to conclude the final result. What I am asking is how does the existence of an $x \in (0,1)$ such that $f'(x) = 0$ mean that there exists a real root between $(0,1)$? All I can conclude from that is that there is an $x \in (0,1)$ such that the slope is the same as a secant line from the endpoints.

StubbornAtom
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D.C. the III
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  • Various solutions: https://math.stackexchange.com/q/200034/42969, https://math.stackexchange.com/q/164818/42969, https://math.stackexchange.com/q/363052/42969, https://math.stackexchange.com/q/448173/42969 – Martin R May 02 '19 at 19:46
  • It is precisely from Baby Rudin that I took the exercise. The solution you posted all discuss using the integral. I was able to arrive at the conclusion, but I don't see any intuition as to why $f'(x) = 0$ means that a root of the original function exists (without the ideas of integration) – D.C. the III May 02 '19 at 19:55
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    But $C_{0} + C_{1}x + ...+ C_{n-1}x^{n-1} + C_{n}x^{n}$ is the derivative, see for example https://math.stackexchange.com/a/164819/42969. – Martin R May 02 '19 at 19:58
  • Set f' = P in the answer. You may write its anti-derivative f. You don't need a chapter on integration to do this. – GNUSupporter 8964民主女神 地下教會 May 02 '19 at 20:01
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    @MartinR I see what you were getting at now. Sorry for the duplicate and thank you for the help. – D.C. the III May 02 '19 at 20:11

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Let $P(x) = C_0 + C_1x + \cdots + C_n x^n$. Then $$ \int_0^1 P(x) \, dx = C_0 + \frac{C_1}{2} + \cdots + \frac{C_n}{n+1} = 0. $$ If $P$ doesn't vanish on $[0,1]$ then since it is continuous, it must be either strictly positive or strictly negative. In the first case, the integral would be strictly positive, and in the second, strictly negative.

Yuval Filmus
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