calculus
$f$ is differentiable in $[a,b]$, and satisfies $f(a)<f(b)$, $f'(a)=f'(b)=0$.
Show that there exists $c$ in $[a,b]$
\begin{equation}
\frac{f(c)-f(a)}{c-a}=f'(c)
\end{equation}
How is the mean value theorem
\begin{equation}
\frac{f(b)-f(a)}{b-a}=f'(c)
\end{equation}
applicable here?
Stuff added at the bottom addressing this question and this other question on the same theme.
Rewritten.
We are going to argue by contradiction. That is: we want to show that there is at most one number $p$ between $-2$ and $2$ where $f(p)$ is equal to $0$. In order to do this, we are going to assume that the opposite of this is true, and from that assumption we are going to deduce something which is absurd or impossible. If we can do this (conclude something which is absurd or impossible), this will mean that our assumption that the conclusion is not true is incorrect; this will mean that the conclusion must be true. This is called a "proof by contradiction".
The opposite of "there is at most one number $r$ between $-2$ and $2$ where $f(x)$ is equal to $0$" is
There are at least two numbers, $a$ and $b$, between $-2$ and $2$, with $f(a)$ and $f(b)$ both equal to $0$.
From this assumption we are going to conclude something absurd/impossible.
So, let us suppose that we have two numbers $a$ and $b$, between $-2$ and $2$, with $a \lt b$, and with $f(a)=0$ and $f(b)=0$. I don't care what $a$ and $b$ are, just that they are between $-2$ and $2$, and are different, and $f$ takes value $0$ there.
Notice that $f(x)$ is continuous on $[-2,2]$ (in fact, continuous everywhere) and is differentiable on $(-2,2)$ (in fact, everywhere). So it is also continuous on $[a,b]$ (whatever $a$ and $b$ are), and differentiable on $(a,b)$. That means that we can apply the Mean Value Theorem to $f$ on $[a,b]$.
Remember that theorem; it says:
Mean Value Theorem. If $f(x)$ is continuous on $[a,b]$ and is differentiable on $(a,b)$, then there exists a point $r$ between $a$ and $b$ (that is, $a\lt r\lt b$) where $$f'(r) = \frac{f(b)-f(a)}{b-a}.$$
(Or perhaps you know the conclusion as: "$f'(c)(b-a) = f(b)-f(a)$"; it's the same thing, since you can get from this to the equation I wrote by dividing through by $b-a$, which is not zero since $a\neq b$).
What does the Mean Value Theorem tell us for our function $f(x)$, given our assumption? Since $f(b)=0$ and $f(a)=0$, the Mean Value Theorem tells us:
There is a point $c$, between $a$ and $b$ (and so between $-2$ and $2$) where $$f'(r) = \frac{f(b)-f(a)}{b-a} = \frac{0-0}{b -a} = \frac{0}{b-a} = 0.$$
That is: if $f(x)$ really has at least two roots in $[-2,2]$, then there has to be a point $r$ where $f'(x)$ is $0$.
(Yes, you can also conclude this from Rolle's Theorem; if you have $f(a)=f(b)$ in the Mean Value Theorem, you get Rolle's Theorem; Rolle's Theorem is a special case of the Mean Value Theorem, that is, something you get by adding assumptions).
So, from assumption that there are at least two points in $[-2,2]$ where $f(x)$ is equal to $0$, we concluded that there has to be at least one point in $(-2,2)$ where the derivative $f'(x)$ is $0$.
Now, what is the derivative? Since $$f(x) = x^3 - 15x + c,$$ where $c$ is a constant, then taking derivatives we get $$f'(x) = 3x^2 - 15 + 0 = 3x^2-15.$$ We can rewrite it a bit to me it easier: $$f'(x) = 3(x^2-5) = 3(x-\sqrt{5})(x+\sqrt{5}).$$ So we know exactly where $f'(x)$ can equal $0$. The only values of $r$ where we have $f'(r) = 0$ are $r=\sqrt{5}$ and $r=-\sqrt{5}$.
But our assumption that there are at least two points $a$ and $b$ in $[-2,2]$ where $f(x)$ is equal to $0$ led us, inexorably and without any "out", to the conclusion that there has to be a point $r$ in $(-2,2)$ where $f'(r)$ is $0$. The problem is: the only points where $f'(x)$ is $0$ are both outside the interval $[-2,2]$ ($\sqrt{5}$ is more than $2$, and $-\sqrt{5}$ is less than $-2$). So $f'(x)$ is never equal to $0$ in $[-2,2]$.
So we have the following two conclusions:
I trust you see that these two things are impossible to satisfy at the same time. We have reached an absurd or impossible conclusion.
Which is good: that's exactly what we wanted. That's how a proof by contradiction works. From our assumption that there are at least two points where $f(x)$ is equal to $0$ on $[-2,2]$, we reached a contradiction. This means that it is false that there are at least two points where $f(x)$ is equal to $0$ on $[-2,2]$.
That means that on $[-2,2]$ there is at most one point where $f(x)$ can be equal to $0$. And this is the conclusion we wanted to reach.
Note that we didn't find the roots of $f(x)$, we only showed that, no matter what $c$ is, so long as it is a constant, $f(x)$ cannot have more than one root in the interval $[-2,2]$.
Added. The point of this problem and other similar problems you've been struggling with is to show you how the Mean Value Theorem gives you information that you can exploit to get information about the function. It is showing you how the different ideas work together to get you more information than might be obvious at first glance.
Part of the reason you are struggling with understanding is that, as you've stated explicitly elsewhere, you don't care. You are not interesting in understanding what these problems are trying to tell you, you are just trying to get the problems done, quickly, in isolation; you just want to know what to write down to get the grade, and that's it.
The problem is that this, and problems like it, are incredibly hard when you try to do them in isolation; they only become clear when you make the connections they are trying to get you to make, and in order to make those connections you need to go beyond just doing the problem and getting the grade.
The Mean Value/Rolle's Theorem tells you that if you know something about a differentiable function between two points, then you have a little bit of information about the derivative of the function somewhere in between. In a sense, the Mean Value Theorem (and its special case, Rolle's Theorem) is actually very vague and the information it gives is very little. Intuitively, it says that if $f$ is differentiable on $[a,b]$, then there is at least one point in between where the instantaneous rate of change (the value of the derivative) is equal to the average rate of change (the slope of the second line that joins $(a,f(a))$ and $(b,f(b))$. It's vague because it doesn't tell you where, or how many times, just that it happens "at least once". And it is very little information because you have to assume you know a fair amount about the function on the entire interval, and it only tells you that "somewhere", there is "at least one point" where the derivative takes a specific value.
As a special case of the Mean Value Theorem you get Rolle's Theorem: if $f$ is differentiable on $[a,b]$, and it 'starts' and 'ends' with the same value ($f(a)=f(b)$), then there has to be at least one point in between where the derivative is $0$ (because the average rate of change is $0$).
Why is this important? Because we are often interested in knowing where a function is equal to $0$. A lot of problems turn on figuing out whether (if not where) a function is equal to $0$. It is used to solve optimization problems, to find points of equilibrium in engineering, and many, many other places.
As a special case of Rolle's Theorem, which means that as a special special case of the Mean Value Theorem, we conclude that if $f$ is differentiable between two points where $f$ is zero, then there has to be at least one point where the derivative is also $0$.
Turns out we can conclude a lot of things from this "little bit of vague information" for differentiable function. For example: if $f$ is differentiable everywhere, and its derivative is never zero, then the function is zero at most once (because between any two zeros of $f$ there would have to be a zero of $f'$). If $f'$ is zero in just one place, then $f$ can be zero in at most two places (because if it were zero at three place, say at $A$, at $B$, and at $C$, then $f'$ would have to be zero at least once between $A$ and $B$, and at least once between $B$ and $C$, but $f'$ can only be zero once, not twice). If $f'$ is zero in exactly 175 places, then $f$ can have at most 176 zeros (because between any two there is at least one zero of $f'$, and there are only 175 of those; so the number of "gaps between consecutive zeros" of $f$ is at most $175$)
For the case at hand, what does this special special case imply? Since we can compute $f'(x)$, and we can verify that $f'(x)$ is never equal to $0$ on $[-2,2]$, then that means that $f(x)$ is zero at most once on $[-2,2]$.
What does it imply on your other question? If we already know that each polynomial of degree 10 has at most 10 roots, then that means that a polynomial $p(x)$ of degree 11 can have at most 11 roots, because its derivative $p'(x)$ is a polynomial of degree 10 and so the derivative has at most 10 roots; so there are at most 10 "spaces between consecutive zeros of $p(x)$". We know a quadratic polynomial has at most 2 roots, so that means that a cubic polynomial can have at most two "spaces between consecutive zeros", so a cubic polynomial can have at most 3 roots.
These are all conclusions we can derive from the Mean Value Theorem; they are all connected to learning how to gain information about $f(x)$ from information we may already have, or can easily find out, about $f'(x)$. In this present problem, we can easily find information about the zeros of $f'(x) = 3x^2 - 15$, and this information in turn tells us things about the zeros of all functions that have $f'(x)$ as their derivative (which are the functions of the form $$f(x) = x^3 - 15x + C,\qquad C\text{ a constant.}$$ Note that we have gained information about an infinite number of different functions, just from knowing a little bit about $f'(x)$. This remarkable feat can be done by understanding what the Mean Value Theorem is telling us, not by "sort of memorizing" (incorrectly, based on your paraphrases that confuse the slope of a line with the line) it.
Suppose to the contrary there are two distinct values such that $f(a)=a$, and $f(b)=b$.
The defining $g(x)=f(x)-x$, we get $g(a)=0$ and $g(b)=0$.
By Mean Value Theorem (Note $g$ is also differentiable):
$g'(c)=\frac{g(b)-g(a)}{b-a}=0$ for some $c$ between $a$ and $b$.
This means $0=g'(c)=f'(c)-1$, so $f'(c)=1$, a contradiction.
Best Answer
Consider the auxiliary function $$g(x):={f(x)-f(a)\over x-a} \quad(a<x\leq b), \quad g(a):=0\ .$$ Then $g$ is continuous on $[a,b]$ and differentiable in $(a,b)$. One computes $$g'(x)={(x-a) f'(x)-\bigl(f(x)-f(a)\bigr)\over (x-a)^2}\qquad(0<x\leq b)\ .$$ If $g(b)=0$, then this is a consequence of Rolle's Theorem.
If $g(b)>0$, then $g'(b)<0$; therefore $g(x)>g(b)>0$ immediately to the left of $b$. This allows us to conclude that $g$ assumes its maximum on $[0,b]$ in an interior point $c$. From $g'(c)=0$ it then follows that $${f(c)-f(a)\over c-a}=f'(c)\ .$$
The case for $g(b)<0$ is similar.