## Rolle’s Theorem and Lagrange’s Mean Value Theorem in Detail

Rolle’s Theorem and Lagrange’s Mean Value Theorem are one of the extensively used theorems in advanced calculus. An Indian mathematician and astronomer Vatasseri Parameshvara Nambudiri introduced the concept of the mean value theorem. Later mean value theorem was proved by Cauchy in 1823.

Later in 1691, Michel Rolle proved a special case of mean value theorem So it was named after him, Rolle’s mean value theorem. In 1834, Moritz Wilhelm Drobisch, a german mathematician used this name first of all, and later in 1846, it was used by Giusto Bellavitis.

In this article, we will understand, Lagrange’s mean value theorem formula, Lagrange’s mean value theorem statement, and proof of Lagrange’s mean value theorem.

## State and Prove Lagrange’s Mean Value Theorem

Mean Value Theorem (MVT): For any given curve between two endpoints, there must be a point at which the slope of the tangent to the curve is the same as the slope of the secant through its endpoints.

Mathematically, The statement of Lagrange's mean value theorem is

If f(x) is a function defined on [a,b] and

(i) f(x) is continuous on [a,b]

(ii) f(x) is differentiable on (a,b)

Then, there exists a real number \[c\epsilon(a,b)\] such that

\[f(c)=\frac{f(b)-f(a)}{b-a}\]

### Lagrange's Mean Value Theorem Proof

To understand the proof of Lagrange’s Mean Value theorem first we need to know the Taylor series. It says if f(x) has continuous derivative up to n+1 order. Then f(x) can be written in the following way

\[f(x)=\sum_{n=0}^{\infty}f^{n}(c)\frac{(x-c)^{n}}{n!}\]

In our problem statement, it is given that the function is continuous and derivable at least one time. By using n = 1 we’ll get

\[f(x)=f(c)+f(c)(x-c)\]

Since f(x) is continuous on [a,b] so f(a) and f(b) must exist. But the point where we need to notice that we don’t know whether \[f’\] is defined on a or not. So we can not say anything about it. Let’s set x = b and d = a substitute this in the previous equation

\[f(b)=f(a)+f’(c)(b-a)\]

Which is nothing but

\[\frac{f(b)-f(a)}{b-a}=f’(c)\]

So, Lagrange mean value theorem formula =\[\frac{f(b)-f(a)}{b-a}=f’(c)\]

Where c is between a and d. That is \[c\epsilon(a,b)\].

Since we learned how to State And Prove Lagrange’s Mean Value Theorem, let us look at the geometrical interpretation of Lagrange’s mean value theorem.

### Geometrical Interpretation of Lagrange’s Mean Value Theorem

Geometrically, Lagrange’s Mean Value Theorem states that If the function is continuous and smooth in some interval then there must be a point (which is mention as c in the theorem) at which the slope of the tangent of the function will be equal to the slope of the secant through its endpoints.

(Image will be uploaded soon)

In the given figure the graph of function f(x).

Let A = (a, f (a)) and B = (b, f (b))

At point c where the tangent passes through the curve is (c, f(c)).

The slope of the tangent line is the same as the secant line i.e both are parallel to each other.

### Rolle’s theorem

Rolle’s theorem is a special case of Lagrange’s mean value theorem. Statement of Rolle’s theorem is, if a function is defined on [a,b] and

(i) f(x) is continuous on [a,b]

(ii) f(x) is differentiable on (a,b)

(iii) f(a) = f(b)

Then there exists a real number \[c\epsilon(a,b)\] such that

\[f'(c)=0\]

Rolle’s mean value theorem proof:

Observe that the first two conditions in Rolle’s theorem are the same as Lagrange’s mean value theorem. The only additional condition is the third condition that is f(a) = f(b).

Using Lagrange’s mean value theorem we have

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

But we have to assume that f(a)=f(b). Using this in the previous equation we get

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

\[f'(c)=0\]

Where \[c\epsilon(a,b)\]

### Geometrical interpretation of Rolle's mean value theorem

(Image will be uploaded soon)

Let us consider f(x) is a continuous curve in [a,b] (including corner points). We also assume that function is smooth (assuming differentiable means assuming smoothness in the curve). Since the function is differentiable we can talk about its tangent. So, the tangent at each point in (a,b) will exist. Now let’s use the last condition which says f(a) = f(b). It means we want the tangent to be parallel to the x-axis and when it does, the first derivative of the function will become zero.

From the algebraic perspective, if f(x) is a polynomial function of any degree (not infinite) and the value of the polynomial at x = a and x = b are equal then there exists at least one root in the interval (a,b) of \[f'(c)=0\].

Here we need to give special attention to “at least one root”. It means theorem is giving assurance for a minimum of one root, it could be more than that (of course depending on the function). So, it can be concluded the converse of Rolle’s theorem is not true.

Consider a cubic equation f(x) = ax3 +bx2 +cx,

If a + b + c = 0, the equation 3ax2 +2bx +c = 0

Now f(0) = 0.

So, x = 0 is the root of the above equation.

Also, f(1) = a + b + c.

But it is given that a + b+ c = 0.

So, x = 1 is also a root of the above equation.

Now f(0) = f(1).

Applying Rolle's Theorem, there exist at least one 0 < c < 1 such that

f’(c) = 0

Hence, (3ax2 +2bx+c)x=c = 0

Hence the quadratic equation

3ax2 + 2bx + c = 0 has at least one real root.

### Solved Example

Question: For the given function f(x) = \[x^{2}-5x+6\] in the interval [a, b], where a = 2 and b = 3, Verify Mean Value Theorem.

Answer: We have given the function \[f(x)=x^{2}-5x+6\]. Since it’s a polynomial of degree two so it’ll be continuous throughout R. Now let’s find its derivative

\[f'(x)=2x-5\]

According to Lagrange’s mean value theorem

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

Here,

a = 2 and b = 3. Now let’s calculate f(2) and f(3).

f(2)=\[2^{2}-5\times2+6\]

\[\Rightarrow f(2)=4-10+6\]

\[\Rightarrow f(2)=0\]

And, f(3)=\[3^{2}-5\times3+6\]

\[\Rightarrow f(3)=9-15+6\]

\[\Rightarrow f(3)=0\]

Now let’s use the mean value theorem

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

\[\Rightarrow 2c-5=\frac{0-0}{3-2}\]

\[\Rightarrow 2c-5=0\]

\[\Rightarrow 2c=5\]

\[\Rightarrow c=\frac{5}{2}\]

Observe that \[\frac{5}{2}\epsilon(2,3)\]

Since \[\frac{5}{2}\epsilon(2,3)\] which verifies the Lagrange’s mean value theorem.

This problem also verifies Rolle’s mean value theorem since f(a) = f(b).

## FAQs on Rolle’s Theorem and Lagrange’s Mean Value Theorem

**1. What are the Major Differences Between Lagrange’s Mean Value Theorem and Rolle’s Theorem?**

Let’s assume, f is a Function and a and b are two variables in closed intervals, continuous, differentiable on an open interval. Then, a tangent is parallel to joining sections, (a, f(b)) and (b, f(b)) at c point in the mean theorem. On the other hand, in Rolle’s Theorem, a tangent is parallel to x-axis at c point.

While the main concept is Lagrange’s mean value theorem, Rolle’s theorem is a special variant of it or extension of the primary concept. This difference helps students to learn about both these concepts well.

**2. Where can you not Use Rolle’s Theorem?**

Even if a function f continuous on a closed interval, differentiates in an open interval and f(a)= f(b), then you can apply Rolle’s theorem to solve a problem. However, there are some exceptions as well that you need to understand where you cannot use this theorem.

For instance, f(x) ={x} (where {x} is a fractional part function) on 0,1, a closed interval . You can derive from the function that on this open interval (0,1) derivative equals 1, everywhere. In such a case, Rolle’s theorem does not apply because f(x), the function has a discontinuity or break at x = 1 which implies the function is not continuous on all places of 0,1, on the closed interval.

**3. What does Rolle’s Theorem Show?**

Rolle’s theorem states that in the case of a constant function, the graph of it would be a horizontal line segment. Simultaneously, it also fulfills all conditions of Rolle’s Theorem as the derivative is 0 everywhere.

However, you need to remember that this theorem guarantees a minimum of one point if not multiple points. Yet, to answer this kind of question, you also have to provide a graphical representation of this theorem to substantiate your answer. Only then, you can secure full marks in this part.

**4. State Lagrange’s mean value theorem.**

Lagrange’s mean value theorem states that if f:[a,b]\[\rightarrow R\]. and f is continuous on [a,b] and differentiable on (a,b) then there exist a point c in (a,b) such that

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

**5. Prove that, The differentiation of a function is zero if and only if it is a constant function using Lagrange’s mean value theorem.**

According to Lagrange’s mean value theorem

\[f(b)=f(a)+f'(c)(b-a)\]

**Case-I**

Consider that differentiation of a function is zero and we’ll prove that the function is constant.

Since the derivative of the function is zero so \[f'(c)\] will be zero in Lagrange’s mean value theorem. Which will give us

\[f(b)=f(a)+0\times(b-a)\]

\[f(b)=f(a)\]

Since c could be any point in between a and b ( an arbitrary point) and the derivative is zero throughout the interval and the value at edge points are the same. Hence the function is constant.

**Case-II**

Conversely, let the function is constant and we’ll prove that its derivative will be zero.

Since the function is constant it means f(a)=f(b) for b>a. Using this in Lagrange’s MVT we get

\[f(b)-f(a)=f'(c)(b-a)\]

\[0=f'(c)(b-a)\]

Product of:\[f'(c)\] and \[(b-a)\] is 0 then either of them has to be zero.\[(b-a)\] can’t be zero because on a number we considered b>a and hence \[f'(c)\] has to be zero. This concludes the fact that the derivative of any constant function is zero.

**6. Given \[f(x)=x^{2}+8x+14\]. Calculate all values of c in the interval \[(-6,-2)\] so that \[f'(c)=0\]**

The function given in the question is a quadratic polynomial. So it is continuous and differentiable everywhere. So, Calculating the values of the function at the endpoints of the given interval

\[f(-6)=(-6)^{2}+8.(-6)+14=36-48+14=2\]

\[f(-2)=(-2)^{2}+8.(-2)+14=4-16+14=2\]

Since both of them are equal to each other we can conclude that all the three conditions of Rolle’s theorem are satisfied. So, we can apply this theorem to find the value of c, differentiating the above equation:

\[f’(x)=(x^{2}+8x+14)’=2x+8\].

Solving the equation and find c:

\[f'(c)=0,\Rightarrow 2c+8=0,\Rightarrow c=-4\].