How does Rolle's Theorem prove that a function is constant?

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In summary, the conversation discusses Rolle's Theorem and its proof using the Extreme Value Theorem. The theorem states that if a function on a closed interval attains both a supremum and an infimum, and if its derivative at a point is 0, then the function must be a constant. The proof involves showing that at the point where the function changes from positive to negative slope, the left and right limits are equal. The conversation also addresses confusion about the two cases presented in the proof, with one involving a positive result and the other a negative result, and explains that this is due to the limit being reached at the top of the function.
  • #1
sponsoredwalk
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Hey PF'ers :biggrin:

I'm having trouble with a proof of Rolle's Theorem.

I understand that by the Extreme Value theorem the function on the closed interval [a,b] must attain both a supremum and an infimum
so if [tex]f(x) = f(a) = f(b) \forall x \in [a,b] and f'(c) = 0[/tex] then it must be a constant function.

That is the trivial case.

If I understand the proof correctly from here it goes as follows.

By the Extreme Value Theorem applied to Rolle's theorem there must exist a point c ∈ [a,b] where the function changes from positive slope to negative slope.

What we want to do is show that at the point f(c) the left and right limits, though both being of opposite sign, will be equal to the same point in the end.

It's just that in the following link, http://www.maths.abdn.ac.uk/%7Eigc/tch/ma2001/notes/node42.html" the proof has me confused.

It's just the two cases they show in Rolle's Proof, the two if h>0, & if h<0, parts I need help understanding.

1: Is the first one saying "if you go beyond the supremum c by a small amount h you'll end up below the value f(c) and because of this you'll end up with a value less than 0 (i.e. be negative) ?

2: Also, in the h<0 part why does that become positive?
If you're at the supremum of any function and you move either way you can only move down!

I might be mixing something up or misunderstanding something about 2:.

I think I may be most confused the the limits they use, they are both coming from the left side & shouldn't one be from the left and the other be from the right?

If This is so, shouldn't the first limit be coming from the right (i.e. from the positive side of the graph toward the negative side)?
 
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  • #2
hi,

"
2: Also, in the h<0 part why does that become positive?
If you're at the supremum of any function and you move either way you can only move down!
"

f(c+h) - f(c) is negative and h is negative, so

f(c+h) - f(c) / h is positive I think that you might be right about the limit sides
 
  • #3
1. If h>0, then f(c+h) - f(c) will be less than zero. This is because f(c+h) must be less than f(c) (as f(c) is the maximum). Because h is positive, dividing by it still gives a negative value.

2. In the second part, f(c+h) - f(c) is again less than zero, but this time you are dividing by negative h, so the result is positive.
 
  • #4
Wow thanks for the quick responses guys.

Yeah I think I understand it now.

It's just confused me to have a positive result when you use h<0 and you're at the top of the function!

As the limit is reached that positive number goes to zero.

The positive result is the slope of the small increment below c that is striving up to it.
 

FAQ: How does Rolle's Theorem prove that a function is constant?

What is the definition of Rolle's Theorem?

Rolle's Theorem states that if a function is continuous on a closed interval and differentiable on the open interval, with the same values for the endpoints, then there exists at least one point within that interval where the derivative of the function is equal to zero.

How is Rolle's Theorem used to prove the existence of a critical point?

Rolle's Theorem is used to prove the existence of a critical point by showing that there exists a point within the interval where the derivative of the function is equal to zero. This indicates that the function has a horizontal tangent at that point, which is a characteristic of a critical point.

What is the difference between Rolle's Theorem and the Mean Value Theorem?

Rolle's Theorem and the Mean Value Theorem are both fundamental theorems in calculus, but they differ in their assumptions. Rolle's Theorem requires that the function be continuous on a closed interval and differentiable on the open interval, while the Mean Value Theorem only requires the function to be continuous on a closed interval and differentiable on the open interval.

Can Rolle's Theorem be used to determine the value of a function at a specific point?

No, Rolle's Theorem cannot be used to determine the value of a function at a specific point. It only guarantees the existence of at least one point within the interval where the derivative is equal to zero.

What are the real-world applications of Rolle's Theorem?

Rolle's Theorem has applications in various fields, such as economics, physics, and engineering. For example, it can be used to find the maximum or minimum values of a function, which is useful in optimizing processes and designing structures.

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