# Rolle's theorem: what's the right statement of the theorem?

In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,

Suppose that f is continuous on a closed interval $I := [a, b]$, that the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$. Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.

Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?

You are right, taking $f(a) = f(b)$ is sufficient.

But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:

Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.

So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.

• If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative. – smci Aug 5 '18 at 23:51
• This could be an answer on its own. – user279515 Aug 6 '18 at 3:55

Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).

Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.

• OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one. – Ingix Aug 4 '18 at 17:32

The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.