2
$\begingroup$

I am suppose to show that the equation $x^3 - 15x + c = 0$ has at most one root in the interval [-2,2]

I have sort of memorized the mean value theorem but I don't really understand how it is applicable to this.

2 Answers 2

15

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:

  1. $f'(x)$ has to be zero somewhere between $-2$ and $2$; and
  2. $f'(x)$ is never zero between $-2$ and $2$.

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.

  • 0
    I don't really understand what this means. I guess what is happening is that you are rewriting the order of the theorem to be set to the derivative which means that there should be zeroes of the derivative but how does that relate to a root? Also it looks like it has two roots.2012-04-06
  • 0
    The Mean Value Theorem says: "there exists $c$ in $(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$". If $f(a)=f(b)=0$, then this says "there exists $c$ in $(a,b)$ such that $f'(c)=0$". I'm not rewriting the order of anything, I'm using the Mean Value Theorem exactly as it is stated. The Mean Value Theorem tells me there is a point where the derivative has to take a particular value which, in *this* situation, happens to be $0$.2012-04-06
  • 1
    I think you are confusing the mean value theorem with rolle's theorem. MVT doesn't say anything about them being equal, just continuous and differential on the interval.2012-04-06
  • 2
    Rolle's Theorem is a special case of the Mean Value Theorem. If you have $f(a)=f(b)$ in the Mean Value Theorem, you get the same conclusion as Rolle's Theorem. I'm not confusing anything.2012-04-06
  • 0
    @Jordan: I think that you need to read your notes/textbook a bit more carefully.2012-04-06
  • 0
    I am looking at my book right now.2012-04-06
  • 0
    I said "more carefully".2012-04-06
  • 0
    Why? What did I say that was incorrect? It wasn't constructive or useful to tell me to do that unless you are just trying to troll.2012-04-06
  • 0
    @Jordan: yes, what you said is incorrect. As Arturo said, Rolle's Theorem is a special case of the MVT (i.e. precisely the case where $f(a)=f(b)=0$). And you don't look like you are making any effort to understand any of the two answers given to you.2012-04-06
  • 0
    @Martin Then why am I even asking questions? I want to understand this so I can pass my test, there is no reason for me not to want to. Also I am confused by the word root, is that an x or y intercept?2012-04-06
  • 0
    Fair enough; and we are here because we are trying to help you. "Root" is an $x$ intercept, i.e. a point where your function is zero. So $a$ is a root of $f$ (one usually uses the word when $f$ is a polynomial) if $f(a)=0$.2012-04-07
  • 0
    @Arturo Maybe this is the wrong way to look at it, but was this problem as simple as just finding the roots of the derivative? I guess I am still a little confused, but I know the key here is that the interval limits it because a function of degree 3 can have 2 roots but the interval will change.2012-04-07
  • 0
    @Jordan: It is only "as simple as finding the roots of the derivative" **if** you make the connection between the roots of the derivative and the roots of the original function, which is to be found in the intermediate value theorem/Rolle's theorem. That's a connection that it seems you have had a very hard time making. So I would say, "no, not as simple as that, because there is some argument that needs to be made and you do not seem to understand that argument."2012-04-07
  • 0
    @Arturo Then would it be fair to say that the theorem states that the function has to have two roots for the derivative to have a root?2012-04-07
  • 0
    @Jordan: No; the implication goes the other way. The Theorem tells you that **if** the function has two roots, then the derivative must have a root "somewhere in between". You can have *no* roots of the function and have the derivative have a root ($f(x) = x^2+1$ has no roots, but its derivative $2x$ has a root).2012-04-07
  • 0
    @Arturo Okay I really do not understand. I must have forgot everything I learned overnight. Mean value theorem just states that there will be a derivative that is equal to the secant lines on an interval. That is still true in this example. Algebraically we can see that there is at most one root in the interval, but I don't quite understand why the application of Mean Value Theorem is even needed. To use MVT theorem we would need two seperate points on the interval that equal zero and then find a zero derivative inbetween.2012-04-07
  • 0
    @Jordan: Mean value Theorem says that there is a point where the tangent is parallel to the secant between endpoints (so the derivative is equal to the **slope** of the secant, not to the secant). You don't *need* the MVT for this example, they are trying to get you to make the connection between zeros of a function and zeros of the derivative. I don't think you forgot anything: I think you have yet to understand it.2012-04-07
  • 0
    @Arturo I do not understand, or at least I do not know how to make the connection to MVT. I know that the number of zeroes can not exceed the degree of the function so that means that the derivative will always be a degree less than the function so the zeroes of the derivative can not exceed the degree of the function which means that the critical numbers can not exceed the degree of the function which means that the function cannot have more zeroes. I do not know how to translate that to the MVT. So the function has at most n zeroes for the n degree.2012-04-09
  • 0
    @Jordan: "Degree" only makes sense for polynomials; not all functions are polynomials. *This* problem is not about the number of zeroes, it's about the connection between the *location* of zeros of a function and zeros of its derivative, and how two zeros of a function forces a zero of the derivative *in a certian location*. **It's not about degrees**. *This* problem is not about degrees of polynomials. Everything you say in your comment is irrelevant to this question.2012-04-09
  • 0
    @Arturo So from the MVT I can say that it is true that for a derivative to have a zero the function must have 2 points that are equal to each other?2012-04-09
  • 0
    Actually that now sounds wrong to me, I think there can be zero derivatives without two equal values on a function (although the points right next the to zero derivative would be equal right?). I think it might be more accurate to say that if a function has to points f(a) and f(b) that are equal than there has to be a derivative between them. But this seems to go the opposite direction to prove zeroes. It seems like I am going to the wrong way.2012-04-09
  • 0
    @Jordan: Yes, that's wrong. You are still trying to go the wrong way. The MVT tells you, among other things, that **if** the function takes the same value at two points (as you write, "if $f(a)$ and $f(b)$ are equal", with $a\neq b$, **then** the derivative must have *at least one zero* in between. But this statement is logically equivalent to the following: if the derivative **does not** have any zeros between $a$ and $b$, then $f(a)$ and $f(b)$ cannot be equal.2012-04-09
  • 0
    Thank you for the comprhensive write up. Are these things I should be learning on my own though as I go through the material? I am not sure where I would pick these kinds of idea up.2012-04-10
  • 1
    @Jordan: These are things that the problems are supposed to be guiding you to recognize/realize, either by solving them yourself, or by getting help in getting them. But their objective is not to simply have you perform this particular problem, but to get the connection and to see the theorems in action.2012-04-10
  • 1
    I usually do all the homework that I am capable of but I never get these deeper meanings out of the problems. This is the second time I have taken the class and I have done all the problems on the MVT but I never even came close to understand it. It feels like there is something else I need to do.2012-04-10
  • 0
    isn't the part you described as-"...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 ..." refer to the Lagrange's Mean Value theorem?2018-06-02
2

What you need here is Rolle's Theorem (which is a particular case of the MVT). This tells you that if $f(x)=x^3-15x+c$ and $f$ has two zeros somewhere (zeroes of $f$ are exactly the roots of your equation) then its derivative has a zero in between.

Now, in this case, $f'(x)=3x^2-15$, which is zero at $-\sqrt5$ and $\sqrt5$. In particular, $f'$ is never zero in the interval $[-2,2]$. So there cannot be more than one zero in the interval $[-2,2]$.

  • 0
    I don't understand how Rolle's is applicable to this, if I have a c how can I know that the 2 functions are equal for the endpoints?2012-04-06
  • 1
    The $c$ in the statement is **not** the $c$ from the Mean Value/Rolle's Theorem. The $c$ in the problem is an arbitrary constant. Change it to $D$ if it is confusing you.2012-04-06
  • 0
    I just meant that it is any number, so if it is 5 won't that change everything? Because that will change the value of the function so I will not even know if Rolle's is applicable. For example f(-2) = 22 + c and f(2) = -22 + c which will not equal eachother.2012-04-06
  • 0
    The $c$ is fixed throughout the problem. Make it 100, if you want. And there are not two functions, just one. And Rolle is always applicable if your function is differentiable and you are considering two zeroes of it (where, by definition, $f(a)=f(b)=0$ is $a,b$ are said zeroes).2012-04-06
  • 0
    My book is telling me that f(a)=f(b).2012-04-06
  • 0
    @Jordan: And if $f(a)$ is $0$, and $f(b)$ is *also* $0$, then is it **not** the case that $f(a)=f(b)$?2012-04-07
  • 0
    @Arturo Yes that is true but it doesn't appear to be true for this problem. f(-2) != f(2)2012-04-07
  • 0
    @Jordan: And we did **not** say that $a$ was -2 and $b$ was $2$. $a$ and $b$ are two points *inside* the interval $[-2,2]$, where we are assuming that $f$ is equal to $0$. We are trying to prove the result by contradiction.2012-04-07
  • 0
    @Arturo But then what points are there then that mean f(a)=f(b)?2012-04-07
  • 0
    @Jordan: Read my edited answer. We are trying to prove that there is at most one place where $f(x)$ is equal to zero by contradiction: we assume there are at least two, and call those two $a$ and $b$. The only things we are assuming are that $a$ and $b$ are between $-2$ and $2$, and $f(x)$ is zero at both. We apply Rolle's Theorem/Mean Value Theorem to $f(x)$ on $[a,b]$ (whatever $a$ and $b$ happen to be), not to $f(x)$ on $[-2,2]$.2012-04-07