Does this variant on Rolle's theorem have a name? The following seems to be true:

Suppose $f : (a,b) \rightarrow \mathbb{R}$ is continuously differentiable. Then $$|x \in (a,b):f(x) = 0| \leq |x\in (a,b):f'(x) = 0|+1.$$
(By $|x \in X : P|$, I just mean the number of $x \in X$ satisfying $P$. This can be viewed as shorthand for the more long-winded $|\{x \in X : P\}|$.)

I'm having trouble formalizing the details, but the proof is basically by Rolle's theorem. Since $f$ is continuous, there are two cases:


*

*The zeroes of $f$ are isolated from each other.

*There exists a non-empty open interval on which $f$ is zero.


In the second case, both the LHS and RHS are $|\mathbb{R}|$, so we're done. So assume the zeros of $f$ are isolated from each other. Then we repeatedly apply Rolle's theorem to get a stationary point of $f$ for each root of $f$ beyond the first. That's the idea.
Anyway, just wondering if this result has a name? It's basically the same as Rolle's theorem, but perhaps bit easier for young people to understand and use. For instance, suppose we wish to prove that $\sqrt{2}$ refers to exactly one real number. Define $f(x)=x^2-2.$ We want to show that $f$ has precisely one root in $(0,\infty)$. We get a lower bound on the cardinality its set of roots by computing $f(0) = -2$ and $f(2) = 2$, hence by IVT, we have:
$$1 \leq |x \in (0,\infty) : x^2 -2 = 0|$$
For an upper bound, use the above theorem to get
$$|x \in (0,\infty) : x^2 -2 = 0| \leq |x \in (0,\infty) : 2x = 0|+1 = 0+1 = 1$$
Ergo: 
$$|x \in (0,\infty) : x^2 = 2| = 1$$
 A: You reach a wrong conclusion. A function can be continuously differentiable and have infinitely many zeros in an open interval, without being constant in any subinterval.
Consider
$$
f(x)=\begin{cases}
0 & \text{if $x=0$} \\[6px]
x^3\sin\dfrac{1}{x} & \text{if $x\ne0$}
\end{cases}
$$
Then $f$ is everywhere differentiable and
$$
f'(x)=\begin{cases}
0 & \text{if $x=0$} \\[6px]
3x^2\sin\dfrac{1}{x}-x\cos\dfrac{1}{x} & \text{if $x\ne0$}
\end{cases}
$$
which is continuous. On the other hand, there is no interval where $f$ is constant.
If the set $Z(f)$ of zeros of $f$ is finite, say $Z(f)=\{x_0,x_1,\dots,x_n\}$, we can define $\varphi\colon Z(f)\setminus\{x_0\}\to Z(f')$ by assigning to $x_i$ an arbitrarily chosen zero of $f'$ in the interval $(x_{i-1},x_i)$, whose existence is guaranteed by Rolle’s theorem. The choice can be made canonical by choosing the leftmost point of absolute maximum in $(x_{i-1},x_i)$ if $f$ assumes positive values in the interval, the leftmost point of absolute minimum otherwise (note that $f$ cannot be constant over the subinterval).
If $Z(f)$ is infinite, the situation is hairier. In the case it is countable, the set contains either an increasing or a decreasing sequence. Rolle’s theorem can be easily applied in the same way as before.
What about the uncountable case?
A: This variant is stated as exercice V.1.12 in the book Problems and Theorems in Analysis (Tome 2) by Polya and Szego, and is there an anonymous early consequence of Rolle's theorem. So I don't think it has a well-known name.
You can state it as
$$
Z(f) \le Z(f')+1,
$$
where $Z$ gives the number of zeros of a function.
Note. Pick up a copy of Problems and Theorems in Analysis if you can. It's an accessible read and the exercices start easy in each section.
