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).
You can use the wavy curve approach to find the range in which the local extremas and global extremas.

The wavy curve method is specifically used for polynomial equations.

Using this method you can easily come to know that there is no global maxima and global minima. However in the range (-8,8) there exists a local maxima. That point is x=-8/7