# Statement about Rolle's Theorem (true or false?)

There's a statement, that I believe is false

Between two distinct zeroes of a polynomial $p$, there is a number $c$ such that $p′(c) = 0$.

Here is my reasoning:

1. A polynomial of an even degree has a derivative of an odd degree, so it has no root, in this case the theorem fails.
2. The statement doesn't say that there's at least a number $c$.

Therefore, the statement fails. Is my thinking process correct?

• So you think $x^2$ has no root? – zhw. Apr 4 '15 at 5:47
• You might want to refine this thought a little bit. Are you discussing $\mathbb{R}$, or $\mathbb{C}$ etc.? Must the zeros be distinct? can you think of a function which has only one root for which $p'(c)=0$? Must you consider polynomials of odd/even degrees as separate cases? Are you discussing a closed and bounded interval, or all of $\mathbb{R}$? – 9301293 Apr 4 '15 at 5:50
• Thanks about that, I meant polynomial of even degree has derivative of odd.. – odysseo Apr 4 '15 at 5:52
• Polynomials of odd degree always have a real root. Polynomials of even degree sometimes don't ( eg. $x^2+1$ ), but if that even degree polynomial is the derivative of a polynomial with distinct zeroes, then it does. – Callus - Reinstate Monica Apr 4 '15 at 5:54
• The statement in yellow is true (thanks to Rolle's theorem). Also, have a look at your reasoning on the polynomial $x^2-1$. – mickep Apr 4 '15 at 5:56

There's a statement, that I believe is false

Between two distinct zeroes of a polynomial $$p$$, there is a number $$c$$ such that $$p′(c) = 0$$

I'll begin by saying, a function which satisfies the hypotheses of Rolle's Theorem is guaranteed its conclusions.

A polynomial of an even degree has a derivative of an odd degree, so it has no root, in this case the theorem fails.

A polynomial of even degree indeed has a derivative with odd degree. However, this does not imply the existence (or lack thereof) of a function's real roots.

Take for example the even function $$f(x) = x^2$$. It has one real root located at $$x = 0$$. It has an odd derivative $$f'(x) = 2x$$.

It has an infinite number of intervals $$[a, b]$$ such that $$f(a) = f(b)$$, all of which satisfy the hypotheses of Rolle's Theorem.

Think about it this way: suppose by contradiction that there is no point between two zeroes of a polynomial s.t the derivative is 0. Since the derivative is continuous, then between the two zeroes of the polynomial the derivative must either be strictly increasing or strictly decreasing.

However, you clearly cannot have a value $a$ such that $f(a)=0$, increase it for a bit, then get a value $b$ with $f(b)=0$. Thus a contradiction has been reached.