TITLE: Does this variant on Rolle's theorem have a name?
QUESTION [11 upvotes]: 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: 
$$|x \in (0,\infty) : x^2 = 2| = 1$$

REPLY [3 votes]: 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?