Rolle's Theorem: I just want to understand when the Rolle's Theorem applies or not.

[FritoTaco]

Hello,

I was taught to verify the hypotheses of the Mean Value Theorem for a function by plugging in the interval and seeing if it satisfies it or not. Although, if the numbers aren't the same, why do we keep solving the problem? Shouldn't we just say there are no numbers to satisfy the problem?

\(\displaystyle f(x)=\dfrac{1}{x-2}\) in interval \(\displaystyle [3, 6]\)

\(\displaystyle f(3)=1\)

\(\displaystyle f(6)=\dfrac{1}{4}\)

They don't equal each other. Nor is the function a polynomial. how can you tell that there is such value of c (constant) that applies?

Note: I don't need help solving the problem itself, I just want to understand when the Rolle's Theorem applies or not. Thank you.

Rolle's Theorem:Let f be a function that satisfies the following three hypotheses:

1. f is continuous on the closed interval [a,b][a,b]
2. f is differentiable on the open interval (a,b)(a,b)
3. f(a)=f(b)f(a)=f(b)

 

[FritoTaco]

Update: Is the only way to find out if it applies to the Theorem is by solving the problem and seeing if the answer is in the interval or not?

 

[JeffM]

Update: Is the only way to find out if it applies to the Theorem is by solving the problem and seeing if the answer is in the interval or not?

You have not said what the problem is. You have not said what "it" refers to.

As as you have pointed out, the conditions required for Rolle's Theorem do not apply to the given function in the given interval so why is that theorem even an issue?

Please give the problem exactly and completely and summarize what you have tried so far.

 

[FritoTaco]

The problem says this:

Find all numbers \(\displaystyle c\) that satisfy the conclusion of the Mean Value Theorem for the function \(\displaystyle f(x)=\dfrac{1}{x-2}\) on the interval \(\displaystyle [3, 6]\)

As as you have pointed out, the conditions required for Rolle's Theorem do not apply to the given function in the given interval so why is that theorem even an issue?

I know it doesn't apply to the given function in the interval. But I don't understand how you get to knowing that using the theorem. How do you know if it applies to 1. f is continuous on the closed interval \(\displaystyle [a, b]\) or 2. f is differentiable on the open interval \(\displaystyle (a, b)\)

I know 3. f(a) = f(b) because I already plugged it in and got 2 different answers. But let's say it does apply to rule 3. How would I know if it applies to the other 2 rules? For rule 2, I learned that all polynomials are differentiable but this function isn't a polynomial. So it doesn't apply to rule 2 and 3 as far as I know. If it doesn't apply, why do we solve the problem to find values for c? (because in my class we solved for the c value, but now I don't understand why we solved for c if the function in the given interval doesn't even apply to the Theorem?)

 

[JeffM]

The problem says this:

Find all numbers \(\displaystyle c\) that satisfy the conclusion of the Mean Value Theorem for the function \(\displaystyle f(x)=\dfrac{1}{x-2}\) on the interval \(\displaystyle [3, 6]\)



I know it doesn't apply to the given function in the interval.

What is "it," Rolle's Theorem? The problem never mentions that theorem. What is its relevance? Do you think the Mean Value Theorem and Rolle's Theorem are the same? They are not.

But I don't understand how you get to knowing that (MEANING ROLLE'S THEOREM DOES NOT APPLY) using the theorem. How do you know if it applies to 1. f is continuous on the closed interval \(\displaystyle [a, b]\) or 2. f is differentiable on the open interval \(\displaystyle (a, b)\)

This is exactly backwards. You are mentioning the conditions that must exist before you can use the theorem. You must show that the conditions apply without using the theorem. I suspect in this problem you are expected to "see" that the given function is continuous and differentiable on the given intervals without proof. But to prove that, if that is what the teacher is expecting, you have to fall back on the definitions of continuity and differentiability without reference to either Rolle's Theorem or the Mean Value Theorem.

I know 3. f(a) = f(b) because I already plugged it in and got 2 different answers.

The fact that you got two different values proves that \(\displaystyle f(a) \ne f(b)\). So you know right away that Rolle's Theorem is not relevant. The first two conditions are required for the Mean Value Theorem to apply, but the third does not. Now, as I previously explained, you cannot use a theorem to show that the relevant requirements of the theorem apply. Either you are allowed to "see" that they apply or you must go back to first principles.

For rule 2, I learned that all polynomials are differentiable but this function isn't a polynomial. So it doesn't apply to rule 2 and 3 as far as I know. If it doesn't apply, why do we solve the problem to find values for c? (because in my class we solved for the c value, but now I don't understand why we solved for c if the function in the given interval doesn't even apply to the Theorem?)

Yes, all polynomials are continuous and differentiable, but that does not mean that all differentiable functions are polynomials. All puppies are dogs, but not all dogs are puppies. There are functions that are not polynomials and are everywhere continuous and differentiable, for example sin(x). There are functions that are not polynomials and are continuous and differentiable over some but not all intervals.

The question then is whether this function is continuous and differentiable over the indicated intervals. If so, the Mean Value Theorem applies to that function over those intervals.

Does this help?

 

[FritoTaco]

Oh they aren't the same that helps a lot actually, thank you.