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Rolle’s Theorem

In calculus, Rolle’s theorem states that if a differentiable function (real-valued) attains equal values at two distinct points then it must have at least one fixed point somewhere between them where the first derivative is zero. Rolle’s theorem is named after Michel Rolle, a French mathematician. Rolle’s Theorem is a special case of the mean value theorem.

Lagrange’s mean value theorem is also termed as the mean value theorem itself or the first mean value theorem. Commonly, the mean is considered as the average of the given values but in the case of integrals, the method of finding the mean value of two different functions is different. In this article let us learn Rolle’s theorem and the mean value of such functions along with their geometrical interpretation.

 1 What is Rolle’s Theorem? 2 Geometric interpretation of Rolle’s Theorem 3 Rolle’s Theorem Proof 4 Solved Examples 5 Practice Questions on Rolle’s Theorem 6 FAQs on Rolle’s Theorem

Table of Contents

## What is Rolle’s Theorem?

Let us understand Lagrange’s mean value theorem in calculus before we study Rolle’s theorem.

### Lagrange’s Mean Value Theorem Statement:

The mean value theorem states that “If a function f is defined on the closed interval [a,b] satisfying the following conditions: i) the function f is continuous on the closed interval [a, b] and ii)the function f is differentiable on the open interval (a, b). Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a)”.

This theorem is also known with the name “first mean value theorem”. A special case of Lagrange’s mean value theorem is Rolle’s Theorem. Let us now understand what is Rolle’s Theorem.

### Rolle’s Theorem Statement:

Rolle’s theorem states that “If a function f is defined in the closed interval [a, b] in such a way that it satisfies the following condition: i) f is continuous on [a, b], ii) f is differentiable on (a, b), and iii) f (a) = f (b), then there exists at least one value of x, let us assume this value to be c, which lies between a and b i.e. (a < c < b ) in such a way that f‘(c) = 0.”

Mathematically, Rolle’s theorem can be stated as: Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers. Then there exists some c in (a, b) such that f′(c) = 0.

## Geometric Interpretation of Rolle’s Theorem

In the given graph, the curve y = f(x) is continuous between x = a and x = b and at every point, within the interval, it is possible to draw a tangent and ordinates corresponding to the abscissa and are equal then there exists at least one tangent to the curve which is parallel to the x-axis. Algebraically, this theorem tells us that if f (x) is representing a polynomial function in x and the two roots of the equation f(x) = 0 are x = a and x = b, then there exists at least one root of the equation f‘(x) = 0 lying between these values. the converse of Rolle’s theorem is not true and it is also possible that there exists more than one value of x, for which the theorem holds good but there is a definite chance of the existence of one such value.

## Rolle’s Theorem Proof

When proving a theorem directly, you start by assuming all of the conditions are satisfied. So, our discussion below relates only to functions

• that is continuous over [a, b],
• that is differentiable (a, b),
• and have f(a) = f(b).

With that in mind, notice that when a function satisfies Rolle’s Theorem, the place where f′(x)=0 occurs at a maximum or a minimum value (i.e., extrema).

How do we know that a function will even have one of these extrema? the Extreme Value Theorem theorem says that if a function is continuous, then it is guaranteed to have both a maximum and a minimum point in the interval.

Now, there are two basic possibilities for our function.

Case 1: the function is constant.
Case 2: the function is not constant.

Let us look into each of these cases in more detail.

Case 1: the function is constant

For a constant function, the graph is a horizontal line segment.

In this case, every point satisfies Rolle’s Theorem since the derivative is zero everywhere. (Remember, Rolle’s Theorem guarantees at least one point. It doesn’t preclude multiple points!)

Case 2: the function is not constant.

Since the function isn’t constant, it must change directions in order to start and end at the same y-value. It means at some point within the interval the function will either have a minimum, a maximum or both. So, now we need to show that at this interior-point the derivative is equal to zero. the rest of the discussion will focus on the cases where the interior extrema is a maximum, but the discussion for a minimum is largely the same.

Possibility 1: Could the maximum occur at a point where f′>0?

No, because if f′>0 we know the function is increasing. But it can’t increase since we are at its maximum point.

Possibility 2: Could the maximum occur at a point where f′<0?

No, because if f′<0 we know that function is decreasing, which means it was larger just a little to the left of where we are now. But we are at the function’s maximum value, so it couldn’t have been larger. Since f′ exists, but isn’t larger than zero, and isn’t smaller than zero, the only possibility that remains is that f′=0. And that’s it! We have shown that the function must have extrema and that at the extrema the derivative must equal zero!

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## Examples on Rolle’s Theorem

1. Example 1: Verify Rolle’s theorem for the function y = x2 + 1, a = –1 and b = 1.

Solution: The function y = x2 + 1, as it is a polynomial function, is continuous in [– 1, 1] and differentiable in (–1, 1). Also,

f(-1) = (-1)2 + 1 = 1 + 1 = 2

f(1) = (1)2 + 1 = 1 + 1 = 2

Thus, f(– 1) = f(1) = 2

Hence, the function f(x) satisfies all conditions of Rolle’s theorem.

Now, f'(x) = 2x

Rolle’s theorem states that there is a point c ∈ (– 2, 2) such that f′(c) = 0.

2c = 0

c = 0, where c = 0 ∈ (–1, 1)

Answer: Hence Rolle’s theorem is verified.

2. Example 2: Verify Rolle’s theorem for the function f(x) = x2 – 4 x + 3 on the interval [1 , 3], and then find the values of x = c such that f ‘(c) = 0.

Solution: f is a polynomial function, therefore is continuous on the interval [1, 3] and is also differentiable on the interval (1, 3).

Now, f(1) = f(3) = 0 and thus function f satisfies all the three conditions of Rolle’s theorem.

Also, there exists at least one value of x = c such that f ‘(c) = 0.

f ‘(x) = 2 x – 4
f ‘(c) = 2 c – 4 = 0

Calculating the value of c:

c = 2, which lies in the interval (1, 3).

Answer: Rolle’s theorem is verified and c = 2.

3. Example 3: Can we apply Rolle’s theorem on the function f(x) = cos x on the interval [0, 2π]?

Solution:

We know that the cosine function is continuous on [0, 2π] and differentiable on (0, 2π). Now,

f(0) = cos 0 = 1

f(2π) = cos 2π = 1

So f(0) = f(2π).

Since all conditions of Rolle’s theorem are satisfied, yes, we can apply Rolle’s theorem for the given function.

Answer: Yes.

## FAQs on Rolle’s Theorem

### What is Rolle’s Theorem?

Rolle’s Theorem states that, if a function f is defined in [a, b] such that

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

then there exists a value c where a < c < b in such a way that f‘(c) = 0.

### What are the Three Conditions of Rolle’s Theorem?

Rolle’s theorem is a special case of the mean-value theorem of differential calculus. All three conditions of Rolle’s theorem are important for the theorem to be true: Condition 1: f(x) is continuous on the closed interval [a,b]; Condition 2: f(x) is differentiable on the open interval (a,b); Condition 3: f(a) = f(b).

### Is Rolle’s Theorem the Same As Mean Value Theorem(MVT)?

Rolle’s theorem is clearly a special case of the MVT in which f is continuous in the closed interval [a, b], and differentiable in the open interval (a, b). Further for Rolle’s theorem there exists an additional condition that states that there exists a point c in the interval (a, b) such that f(a) = f(a).

### What is Lagrange’s Theorem?

Lagrange’s Mean Value Theorem (first mean value theorem) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a). A special case of Lagrange’s mean value theorem is Rolle’s Theorem.

### What is the Real-Life Application of Rolle’s Theorem?

Rolle’s theorem finds wide use in Physics, astronomy, etc. A real-life example of Rolle’s Theorem: When you throw a ball vertically up, its initial displacement is zero (f(a)=0) and when you catch it again its displacement is zero (f(b)=0). As displacement function satisfy criteria of Rolle’s theorem of continuity and differentiability over (a,b) interval. There exists at least one f′(c)=0, and in fact velocity is zero at the highest point. Here f(x) is the height of the ball and f′(x) is the vertical velocity of the ball. Here we refer x as the maximum point and the f'(x) as the minimum value.

### What is C in Rolle’s Theorem?

Rolle’s Theorem states that if a function f is continuous on the closed interval [a,b], differentiable on the open interval (a,b), and if f(a) = f(b) then there exists a point c in the interval (a,b) such that f'(c) is equal to the function’s average rate of change over [a,b], which is f'(c) = 0.

### Is the Converse of Rolle’s Theorem True?

The converse of Rolle’s theorem is not true. There can be a function f(x) that does not satisfy any of the conditions of Rolle’s theorem and yet f'(x) is 0, for each x belonging to (0, 1).

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