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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.

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  • $\begingroup$ Let $f(x) = x^3/3 - 4x + 7,000,000,000$. Then $f'(x) = x^2 - 4$ has two roots at 2 and -2. Does $f(x) = x^3/3 - 4x + 7,000,000,000$ have any roots between $2$ and $-2$? $\endgroup$
    – fleablood
    Jul 5 '16 at 17:11
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    $\begingroup$ If you've realized the error in your original proposition, I'd suggest rewriting your post and question to reflect that. You don't have to (and shouldn't) delete you first premise but you should indicate you now realize it was false and have a new question. $\endgroup$
    – fleablood
    Jul 5 '16 at 17:13
  • $\begingroup$ "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. " Not merely $c \in [a,b]$ but $c \in (a,b)$ which means ... you are done. Nitpick: c can be maximum or minimum. $\endgroup$
    – fleablood
    Jul 5 '16 at 18:55
  • $\begingroup$ So you've chosen a title that says almost the opposite of what it should say? $\endgroup$
    – zhw.
    Jul 5 '16 at 19:09
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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.

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Hint:

$f$ is necessarily monotonic between any two consecutive roots of $f'$.

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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.

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  • $\begingroup$ wait, but the premise was not about $f'$ having roots between the interval, but $f$ $\endgroup$ Jul 5 '16 at 21:03
  • $\begingroup$ Um.... yeah and that's what I answered. $\endgroup$
    – fleablood
    Jul 5 '16 at 21:05
  • $\begingroup$ If there are two roots of f in an interval then there is at least one root of f' inside the interval. So if there are no roots of f' in the interval there can be at most one root of f. QED $\endgroup$
    – fleablood
    Jul 5 '16 at 21:07
  • $\begingroup$ Two roots of f => one root of f' => contradiction to premise. $\endgroup$
    – fleablood
    Jul 5 '16 at 21:10
  • $\begingroup$ The question was about f having roots. The premise/hypothesis was about f'. It really was. $\endgroup$
    – fleablood
    Jul 5 '16 at 21:12

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