Rolle's Theorem

Q: What is Rolle's theorem?

Rolle's Theorem is a special case of the Mean Value Theorem that states that if a function is continuous over the closed interval [ a , b ], differentiable over the open interval ( a , b ), and f(a) = f(b) , then there exists at least one number c in ( a , b ) such that f'(c) = 0.

Q: Is Rolle's theorem the same as MVT?

Essentially, Rolle's Theorem is the same as MVT. It is a special case of the MVT where f(b ) - f(a) = 0.

Q: What is an example of Rolle's theorem?

An example of Rolle's Theorem is the function f(x) = cos(x) + 2 over the interval [0, 2pi]. Rolle's Theorem states that because this function meets the theorem's requirements, there exists at least one value c such that f'(c) = 0.

Q: How do you prove Rolle's theorem?

Assume that the requirements of Rolle's Theorem hold for a function f . We can prove Rolle's Theorem by considering the two cases: the function is a constant value and the function is not a constant value. If the function is a constant value, then f(a) = f(b) everywhere and Rolle's Theorem applies over the entire interval ( a , b ). If it is not a constant value, then we know the function must change direction in order to start and end at the same function value. So, somewhere inside the graph, the function will either have a minimum, a maximum, or both. Minimum and maximum values occur when f'(x) = 0.

Q: Does Rolle's theorem include endpoints?

Rolle's Theorem states that the value c where f'(c) = 0 is in the open interval ( a , b ). Thus, endpoints are not included.

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There are some theorems or ideas in Calculus that may seem rather obvious. Rolle's Theorem is one such theorem. Let's say you leave your house to go for a walk. After your walk, you return home. Rolle's Theorem says that because you started and ended at the same place, you must have made a turn at some point during your walk. Though this fact seems evident, Rolle's Theorem is a significant discovery in Calculus.

Three Hypotheses / Conditions of Rolle's Theorem

To be able to use Rolle's Theorem, a few conditions must be met. The function should be:

continuous on the closed interval $[a, b]$
differentiable on the open interval height="21" id="2942299" $(a, b)$
height="21" id="2942301" $f (a) = f (b)$

Rolle's Theorem Definition

Now that we've gone over the conditions for Rolle's Theorem, let's look at what this theorem says.

Rolle's Theorem states that if a function $f$ is:

continuous on the closed interval $[a, b]$
differentiable on the open interval $(a, b)$
$f (a) = f (b)$

then there exists at least one number $c$ in $(a, b)$ such that $f' (c) = 0$ .

Geometrically speaking, if a function meets the requirements listed above, then there is a point $c$ on the function where the slope of the tangent line is 0 (the tangent line is horizontal).

Rolle's Theorem slope of the tangent line geometric explanation StudySmarter A continuous and differentiable function f that has points a and b such that f(a) = f(b) has at least one point c where the slope of the tangent line is 0 - StudySmarter Original

In our walking example, Rolle's Theorem says that since we started and ended at the same place, there must have been a movement where we made a turn (the derivative is 0).

Rolle's Theorem vs. The Mean Value Theorem

Recall the Mean Value Theorem, which states that if a function $f$ is:

continuous on the open interval $(a, b)$
differentiable on the closed interval $[a, b]$

then there is a number c such that $a < c < b$ and

$f' (c) = \frac{f (b) - f (a)}{b - a}$

Rolle's Theorem is a "special case" of the Mean Value Theorem. Rolle's Theorem says that if the requirements are met and there are points a and b such that $f (b) = f (a)$ , or $f (b) - f (a) = 0$ , then there is a point $c$ where $f' (c) = 0$ . If we plug in $f (b) - f (a) = 0$ to the Mean Value Theorem equation for $f' (c)$ , we get $f' (c)$ . So, Rolle's Theorem is the case of the Mean Value Theorem where $f (b) = f (a)$ .

Rolle's Theorem Proof

Let's assume that a function f is continuous on the interval [a, b], differentiable on the interval $(a, b)$ , and $f (a) = f (b)$ . Thus, the requirements of Rolle's Theorem are met. We must prove that the function $f$ has a point $c$ where $f' (c) = 0$ . In other words, the point where $f' (c) = 0$ occurs is either a maximum or minimum value (extrema) on the interval.

We know that our function $f$ will have extrema per the Extreme Value Theorem, which says that if a function is continuous, it is guaranteed to have a maximum value and a minimum value on the interval.

There are two cases:

The function is a constant value (a horizontal line segment).
The function is not a constant value.

Case 1: The function is a constant value

Rolle's Theorem proof f(a) = f(b) everywhere StudySmarter This function, which meets the requirements of Rolle's Theorem, has a derivative of 0 everywhere - StudySmarter Original

Every point on the function meets the Rolle's Theorem requirements as $f' (c) = 0$ everywhere.

Case 2: The function is not a constant value

Because the function is not a constant value, it must change direction to start and end at the same function value. So, somewhere inside the graph, the function will either have a minimum, a maximum, or both.

Rolle's Theorem proof geometric explanation StudySmarter This function, which meets the requirements of Rolle's Theorem, has both a minimum and maximum - StudySmarter Original

We must prove that the minimum or maximum (or both) occur when the derivative equals 0.

Extrema cannot occur when $f' (x) > 0$ because when $f' (x) > 0$ , the function is increasing. At an extrema value, the function cannot be increasing. At a maximum point, the function cannot be increasing because we are already at the maximum value. At a minimum point, the function cannot be increasing because the function was a little smaller to the left of where we are now. Since we're at the minimum value, $f' (x)$ cannot be any smaller than it is now.

Extrema cannot occur when $f' (x) < 0$ because when $f' (x) < 0$ , the function is decreasing. At an extrema value, the function cannot be decreasing. At a maximum point, the function cannot be increasing because $f' (x) < 0$ which means $f' (x)$ was larger a little to the left of where we are now. Since we're at the maximum value, $f' (x)$ cannot be any larger than it is now. At a minimum point, the function cannot be decreasing because we are already at the minimum value.

Since $f' (x)$ isn't less than 0 or greater than 0, $f' (x)$ must equal 0.

Rolle's Theorem Step-by-Step Procedure

While no explicit formula is associated with Rolle's Theorem, there is a step-by-step process to find the point $c$ .

1. ensure that the function meets Rolle's Theorem: continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ .

2. plug a and b into the function to guarantee that $f (a) = f (b)$ .

3. If the function meets all requirements of Rolle's Theorem, then we know that we are guaranteed at least one point $c$ where $f' (c) = 0$ .

4. To find $c$ , we can set the first derivative equal to 0 and solve for $x$ .

Rolle's Theorem Examples

Example 1

Show through Rolle's Theorem that $f (x) = \cos (x) + 2$ over $[0, 2 π]$ has at least one value $c$ such that $f' (c) = 0$ . Then, find the maximum or minimum value of the function over the interval.

Step 1: Ensure that f(x) meets the Rolle's Theorem requirements

By nature, we know that the cosine function is continuous and differentiable everywhere.

Step 2: Check that f(a) = f(b)

Plugging in 0 and $2 π$ into $f (x)$

$\cos (0) + 2 = \cos (2 π) + 2 1 + 2 = 1 + 2 3 = 3$

Since

f (a) = f (b) = 3

, we can apply Rolle's Theorem.

Step 3: Set f'(x) = 0 to solve for x

By Rolle's Theorem, we are guaranteed at least one point $c$ where $f' (c) = 0$ . So we can find $f' (x)$ and set it equal to 0.

$f' (x) = - \sin (x) = 0 \sin (x) = 0$

Using our knowledge of trigonometry and the unit circle, we know the the sine function equals 0 when $x = 0$ and multiples of $π$ . However, the only multiples of $π$ within our interval are $π$ and $2 π$ . So, in our interval, $\sin (x) = 0$ when $x = 0, π, 2 π$ .

Step 4: Plug in c values to f(x) to find the maximum or minimum function values

$f (0) = \cos (0) + 2 = 1 + 2 = 3$

$f (π) = \cos (π) + 2 = - 1 + 2 = 1$

$f (2 π) = \cos (2 π) + 2 = 1 + 2 = 3$

$f (x)$ has a maximum value of 3 at $x = 0, 2 π$ and a minimum value of 1 at $x = π$

Example 2

Let $f (x) = x^{3} - x$ . Does Rolle's Theorem guarantee a value $c$ where $f' (c) = 0$ over the interval $[- 1, 1]$ ? Explain why or why not.

To check if we can apply Rolle's Theorem, we must ensure that the requirements are met.

Step 1: Check if f(x) is continuous and differentiable

We know that $f (x)$ is continuous over the given interval because it is a polynomial. We also know that $f (x)$ is differentiable over the interval: $f' (x) = 3 x^{2} - 1$

Step 2: Check if f(-1) = f(1)

When we plug in $f (- 1)$ , we get $f (- 1) = - 1^{3} - (- 1) = 0$ . When we plug in $f (1)$ , we get $f (1) = 1^{3} - 1 = 0$ .

Step 3: Apply Rolle's Theorem

Since, $f (x)$ is continuous over $[- 1, 1]$ , differentiable over $(- 1, 1)$ , and $f (- 1) = f (1) = 0$ , then Rolle's Theorem tells us that there exists a number $c$ such that $f' (c) = 0$ .

Rolle's Theorem - Key takeaways

Rolle's Theorem is a special case of the Mean Value Theorem where $f (b) - f (a) = 0$
Rolle's Theorem states that if a function $f$ is:
- continuous on the closed interval $[a, b]$
- differentiable on the open interval $(a, b)$
- $f (a) = f (b)$ then there exists at least one number $c$ in $(a, b)$ such that f'(c) = 0
To find apply Rolle's Theorem:
- Ensure that the requirements are met
- Check that the endpoints have the same function value
- Set the first derivative of the function equal to 0 and solve for $x$

Frequently Asked Questions about Rolle's Theorem

Rolle's Theorem is a special case of the Mean Value Theorem that states that if a function is continuous over the closed interval [a, b], differentiable over the open interval (a, b), and f(a) = f(b), then there exists at least one number c in (a, b) such that f'(c) = 0.

Essentially, Rolle's Theorem is the same as MVT. It is a special case of the MVT where f(b) - f(a) = 0.

An example of Rolle's Theorem is the function f(x) = cos(x) + 2 over the interval [0, 2pi]. Rolle's Theorem states that because this function meets the theorem's requirements, there exists at least one value c such that f'(c) = 0.

Assume that the requirements of Rolle's Theorem hold for a function f. We can prove Rolle's Theorem by considering the two cases: the function is a constant value and the function is not a constant value. If the function is a constant value, then f(a) = f(b) everywhere and Rolle's Theorem applies over the entire interval (a, b). If it is not a constant value, then we know the function must change direction in order to start and end at the same function value. So, somewhere inside the graph, the function will either have a minimum, a maximum, or both. Minimum and maximum values occur when f'(x) = 0.

Rolle's Theorem states that the value c where f'(c) = 0 is in the open interval (a, b). Thus, endpoints are not included.

Flashcards in Rolle's Theorem3 Start learning

State Rolle's Theorem.

Rolle's Theorem says that if a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there is at least one value c where f'(c) = 0.

Rolle's Theorem is a special case of the Mean Value Theorem where...

f(b) - f(a) = 0 or f(b) = f(a)

How can we interpret Rolle's Theorem geometrically?

If a function meets the requirements of Rolle's Theorem, then there is a point on the function between the endpoints where the tangent line is horizontal, or the slope of the tangent line is 0.

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Rolle's Theorem

Three Hypotheses / Conditions of Rolle's Theorem

Rolle's Theorem Definition

Rolle's Theorem vs. The Mean Value Theorem