Question: Prove the following without using Rolle's Theorem:

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> If $f$ is a polynomial, $f'(a) = 0 = f'(b)$ for some $a < b$, and there is no $c \in (a,b)$ such that $f'(c) = 0$, then there is at most one root of $f$ in $(a,b)$.
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I've already proven this by contraction by assuming that there is more than two roots and showing that it contradicts Rolle's Theorem. Now i'm wondering how I could prove this without using Rolle's Theorem.

Answer: 
$p(x)= (8+x)^3(8-x)^4$ is a polynomial of degree $7$,

with leading term $x^7$.

1)No abs. maximum ,no abs. minimum . $(x \rightarrow \pm \infty )$

2) $3$ zeroes at $x=-8$, $4$ zeroes at $x=8$.

3)$ -8 <x < 8$ : $p(x) >0$.
(Has a relative maximum).
We have that

$$g(x)=(8+x)^3(8-x)^4\implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$

$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$

which is true when

* $x=\pm 8$
* $3(8-x)=4(8+x) \implies 7x=-8 \implies x=-\frac87$

and the latter is a local maximum.