Using Rolle's theorem to prove at most one root

[kevinnn]

Homework Statement

show that the equation x^3-15x+C=0 has at most one root on the interval [-2,2]

Homework Equations

The Attempt at a Solution

I know I need to use Rolle's theorem but I'm not sure how to find the answer. Thanks.

 

[LCKurtz]

Homework Statement

show that the equation x^3-15x+C=0 has at most one root on the interval [-2,2]

Homework Equations

The Attempt at a Solution

I know I need to use Rolle's theorem but I'm not sure how to find the answer. Thanks.

Well, what if it had two roots on that interval?

 

[kevinnn]

That is exactly what the homework tip said and I don't get and see the importance of that yet.

 

[kevinnn]

Ohhh well if it had two roots then the function would have both positive and negative values on the interval.

 

[Dick]

Ohhh well if it had two roots then the function would have both positive and negative values on the interval.

Not necessarily. If the function has two roots then f(a)=f(b)=0 for a and b in [2,-2]. What would Rolle's theorem then tell you?

 

[kevinnn]

That the function is a constant over the interval [a,b]

 

[kevinnn]

No sorry. It tells me that a number c exists in the interval such that f'(c)=0

 

[LCKurtz]

No sorry. It tells me that a number c exists in the interval such that f'(c)=0

And do you see a problem with that?

 

[kevinnn]

Well if f(a)=f(b)=0 then the function is a constant over [a,b] so there are infinitely many points where the derivative equals zero. The opposite of what I was trying to show. So what am I missing.

 

[LCKurtz]

Well if f(a)=f(b)=0 then the function is a constant over [a,b] so there are infinitely many points where the derivative equals zero. The opposite of what I was trying to show. So what am I missing.

Why do you say that? $f(x) = \sin(x)=0$ for $x = n\pi$, but it isn't constant. Being zero at two points doesn't mean identically zero; surely you know better.

You have noted that $f'(c)$ must equal zero on the interval. Take the derivative and look at it and see what you think.

 

[kevinnn]

I did that and when I take the derivative and set it equal to zero I get plus/minus the square root of 5. Which is not even the interval. That would tell me that there are not at most one, but no roots in [-2,2].

 

[LCKurtz]

I did that and when I take the derivative and set it equal to zero I get plus/minus the square root of 5. Which is not even the interval. That would tell me that there are not at most one, but no roots in [-2,2].

No roots of what? Are you talking about $f'$? The problem is asking about $f(x)$.

 

[kevinnn]

You have noted that $f'(c)$ must equal zero on the interval. Take the derivative and look at it and see what you think.[/QUOTE]

I took the derivative. I don't really see how the derivative helps us in this case. It does not help me find roots does it. Not that I know of?

 

[LCKurtz]

To clear away all the mistakes, read post #2, #7, and #11. See if that doesn't help you see it.

 

[LCKurtz]

I did that and when I take the derivative and set it equal to zero I get plus/minus the square root of 5. Which is not even the interval. That would tell me that there are not at most one, but no roots in [-2,2].

No roots of what?

And answer that question.

 

[kevinnn]

Ohhhhh now I got you. So since the maximum and minimum of the graph are at plus minus the square root of five that means that the graph can only be either increasing or decreasing on our interval so the most the graph can cross the x-axis is once. Thanks.

 

[LCKurtz]

Ohhhhh now I got you. So since the maximum and minimum of the graph are at plus minus the square root of five that means that the graph can only be either increasing or decreasing on our interval so the most the graph can cross the x-axis is once. Thanks.

That is correct, but you are using other theorems you don't need. A simpler argument:

1. Suppose f(x) has two roots on the interval.
2. Then by Rolle's theorem f'(c) = 0 for some c on the interval.
3. But f'(c) doesn't equal zero for any c on that interval.

Therefore f(x) doesn't have two roots on the interval.

 

[kevinnn]

Great. Thanks.