Between $2$ consecutive roots of $f'$, there is at least one root of $f$ Prove that between $2$ consecutive roots of $f'$, there is at least one root of $f$.
I understand that a root of $f'$ represents an extreme point. But, for example, $f(x) = \sin(x)+2$ has no roots, but its derivative, $\cos(x)$, has lots of consecutive roots. 
Ok, while I was writing this, I realized that its no "at least" but "there is at most" one root of $f$.
So, I understand that, between $2$ consecutive maximum points, for example, there can be one root, but if that function tries to come back and make another root between the two max poits, it's gotta create a local maximum point between them. But how do I write this mathematically?
Let me try:
By Rolle's, between $2$ consecutive roots $f(a) = f(b)$, there must be a point $c\in [a,b]$ where $f'(c) = 0$, which is a maximum point. 
Or maybe, can I say the following:
between two roots of $f'(x)$, let's say, $f'(m) = f'(n)$ by rolles theorem we have:
$$\exists c\in [m,n] / f''(c) = 0$$
so there's a maximum point between the roots, but I don't know how to prove that this is the only maximum point, and that this maximum point leads to only one root.
 A: Hint:
$f$ is necessarily monotonic between any two consecutive roots of $f'$.
A: Silly me.  I ignored Rolle's theorem (and kind of proved it from scratch-- [confession-- I can't remember the names and conditions of any of these thereoms.  I just remember the intermediate value theoreom and derive them all from common sense when needed]).
"By Rolle's, between 2 consecutive roots f(a)=f(b), there must be a point c∈[a,b] where f′(c)=0, which is a maximum point. "
And that's it!  You've done it.  If $f'(a) = f'(b) = 0$ and $f(x) = f(y) = 0$ for $a \le x < y \le< b$ then there is a $f'(c) = 0$ with $a \le x < c < y \le b$ which contradicts your premise that $f'$ has no roots between $a$ and $b$.
===  old answer ===
"but if that function tries to come back and make another root between the two max poits, it's gotta create a local maximum point between them. But how do I write this mathematically?"
Well, you first consider $f'$ by itself without regard to $f$ and interpret the intermediate value thereom.
$f'(a) = 0$; $f'(b) = 0$ (wolog $a < b$).  And for no $x \in (a,b)$ does $f'(x) = 0$.  Suppose there is an $x$ where $f'(x) > 0$ and and $f'(y) < 0$ with $x,y \in (a,b)$.  Then by IMT there is an $f'(c) = 0$ with $c \in (\min(x,y), \max(x,y) \subset (a,b)$.  This is a contradiction.  So either $f'(x) > 0; \forall x \in (a,b)$ or $f'(x) < 0; \forall x \in (a,b)$.
Now we think about what that means for $f$.  It means $f$ is monotonically increasing or decreasing.  Wolog let's assume $f$ is monotonically increasing.
If $f(x) = 0$ for $a < x < b$ then $f(w) < 0$ for all $a < w < x$ and $f(z) > 0$ for all $x < z < b$.  So there $x$ would be the root between $a$ and $b$.
If $f(x) = 0$ for $a < x <b$... which need not be the case.
A: Suppose that $f'(a)=f'(b)=0$ with $a<b$ and that $f'\not = 0$ in $]a,b[$. If $f$ has two distinct roots $x$ and $y$ in $]a,b[$, then, by Rolle's theorem (since $f(x)=f(y)=0$), there is $c$ in $[x,y]$ such that $f'(c)=0$. Contradiction. 
