Lagrange’s Mean Value Theorem (LMVT theorem)
Rolle's Theorem Class 12 is a variant of the mean value theorem that meets specific requirements. Lagrange's mean value theorem is both the mean value theorem and the first mean value theorem at the same time. In general, the mean can be defined as the average of a set of values. The procedure of determining the mean value of two separate functions is different in the case of integrals. Let's look at Rolle's theorem, as well as the mean value and geometrical meaning of such functions.
State and Prove Lagrange’s Mean Value Theorem (LMVT)
Lagrange’s Mean Value Theorem or First Mean Value theorem states that a function f is defined in the closed interval (a, b), it agrees with the following conditions:
The function f is always continuous in the closed interval (a, b)
The function is always differentiable in the open interval (a, b)
And there includes a value x = c in such a manner that,
f(b)–f(a)f(b)–f(a) / (b-a) = f’ (c)
LMVT Theorem Proof
Consider the following related function,
F (x) =f (x) +λ x
We selected a number λ such that the condition F(a) =F(b) agreed. Then,
F (a) +λ a =f (b) +λb,
= f(b) -f(a) = λ (a-b)
λ = - f(b) -f(a) /b-a
Therefore, we have
F(x) = f (x) - f(b) -f(a)/ b-a (x)
The function f is continuous in the closed interval (a, b), differentiable in the open interval (a, b), and holds equal values at the endpoints of the interval. Hence, it satisfies all the conditions of Rolle’s theorem; there includes a point c in the interval (a,b) such that
F’(c) = 0
It follows that,
f’c - f (b)-f (a) /b-a = 0
Or,
f(b) -f(a) =f’(c) (b-a)
Geometrical Representation of Lagrange’s Mean Value Theorem
The graph given below the curve y = f(x) is in the continuous form, and x-=a and x=b are also differentiable within the closed interval (a, b), then,
According to the Lagrange Mean Value Theorem,
Any function that is continuous on (a,b) and also differentiable on (a,b) has some in the interval (a,b) such that the secant joining the interval's ends is parallel to the tangent at point C.
f ’(c) is equivalent to f (b) –f (a)/b-a
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Solved Example
1. Verify mean value theorem , if
f (x) = x2-4x -3 in the interval a,ba,b where a =1 and b= 4
Solution: f (x) = x2-4x-3, the given variable is continuous in the interval 1,41,4 and derivable (1,4).
The given polynomial equation is holding all the conditions of Rolle’s Theorem.
The f ’(x) = 2x-4
f ’(x) 2c-4
f (4) = 16-16-3 = -3
f (1) = 1-4-3 =6
There included a value of C such that,
f ’(x) = f (b) –f(a)/ b-a = 2c-4 = -3-(6)/4-1 =3/3=1
Therefore, 2C = 4 +1 or C= 5/2€ (1, 4)
State and Prove Rolle’s Theorem
Statement of Rolle's Theorem
Rolle's Theorem is a specific example of Lagrange's mean value theorem, which states:
If a function f is defined in the closed interval [a, b] in such a way that it meets the conditions below.
On the closed interval [a, b], the function f is continuous.
On the open interval, the function f is differentiable (a, b)
If f (a) = f (b), then at least one value of x exists; let us assume that this value is c, which is between a and b, i.e. (a c b) in such a way that f'(c) = 0.
If a function is continuous on the closed interval [a, b] and differentiable on the open interval [a, b], there exists a point x = c in (a, b) where f'(c) = 0. (a, b).
Rolle's theorem can be expressed mathematically as:
Let f: [a, b] R be continuous on [a, b] and differentiable on (a, b), with f(a) = f(b). Then some c in (a, b) exists such that f′(c) = 0.
Rolle's Theorem Geometric Explanation
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If the curve y = f(x) is continuous between x = a and x = b, and it is possible to draw a tangent at any point within the interval, and the ordinates corresponding to the abscissa and are identical, then there is at least one tangent to the curve that is parallel to the x-axis.
This theorem states that if f (x) represents a polynomial function in x and the two roots of the equation f(x) = 0 are x = a and x = b, then at least one root of the equation f'(x) = 0 is between these two values.
The inverse of Rolle's theorem is not true; it is possible that more than one value of x exists for which the theorem is true, but there is a reasonable possibility that only one such value occurs.
Examples of Rolle's Theorem
1. Verify Rolle's theorem using the functions y = x2 + 2, a = –2, and b = 2 using the functions y = x2 + 2, a = –2, and b = 2.Ans. The function y = x2 + 2 is continuous in [– 2, 2] and differentiable in (– 2, 2), according to Rolle's theorem.
From the given question,
f(x) = x(2) + 2
f(-2) = (-2)2 + 2 = 4 + 2 = 6
f(2) = (2)2 + 2 = 4 + 2= 6
Thus, f(– 2) = f( 2)
Hence, the value of f(x) at –2 and 2 coincide.
Now, f'(x) = 2x
Rolle’s theorem states that there is a point c ∈ (– 2, 2) such that f′(c) = 0.
At c = 0, f′(c) = 2(0) = 0, where c = 0 ∈ (– 2, 2).
2. Discuss the conditions for the application of Rolle’s Theorem for the following function: F(x) = x2/3 on −1,1,−1,1
Ans. f ’(x) = 2/3x1/3
f ’ (0) = 2 /3(0)1/3
f ’(0) = ∞
So f ‘(x) does not takes place at x=o € (-1, 1)
f ’(x) is not differentiable in X € (-1, 1)
So, Rolle’s Theorem is not applicable on F(x) in −1,1,−1,1
Facts About Rolle’s Theorem
Rolle’s Theorem was initially proven in 1691.
Rolle’s Theorem was proved just after the first paper including calculus was introduced.
Michel Rolle was the first famous Mathematician who was alive when Calculus was first introduced by Newton and Leibnitz.
Initially, Michel Rolle was critical of calculus, but later he decided to prove this significant theorem.
Quiz Time
1. In which of the following intervals is f(x) = -x satisfy Rolle’s Theorem?
a. (0,2)
b. (3,4)
c. (-3,-1)
d. None of these
2. Based on Rolle's theorem for a continuous function f (x), if the starting point f(a) and the end-point f(b)equals 0 then,
a. It can be somewhere between f (a) and f(b) and the instantaneous rate of change should be 0.
b. It can be somewhere between f (a) and f (b) and the function must be equal to 0.
c. The function is flat
d. Rolle’s Theorem cannot be applied.
3. Consider the function f (x) given below. Based on Rolle’s Theorem between x= o and x= pi, for how many values of x is the instantaneous rate of change equals 0.
a. At least 1
b. Rolle’s theorem is not applicable
c. 2
d. 0
FAQs on Rolle's Theorem
1. What is the formal statement of Rolle's Theorem in calculus?
Rolle's Theorem states that if a real-valued function f is defined on a closed interval [a, b] and satisfies three specific conditions, then there exists at least one number 'c' in the open interval (a, b) such that the derivative of the function at that point is zero, i.e., f'(c) = 0. It is a fundamental result in differential calculus and a special case of the Mean Value Theorem.
2. What are the three essential conditions for Rolle's Theorem to be applicable to a function?
For Rolle's Theorem to apply to a function f(x) on an interval [a, b], the following three conditions must be met as per the CBSE Class 12 syllabus for 2025-26:
- The function f(x) must be continuous on the closed interval [a, b].
- The function f(x) must be differentiable on the open interval (a, b).
- The values of the function at the endpoints of the interval must be equal, i.e., f(a) = f(b).
3. What is the geometric interpretation of Rolle's Theorem?
Geometrically, Rolle's Theorem means that if a curve representing a function y = f(x) is continuous from point A (where x=a) to point B (where x=b), has a unique tangent at every point in between, and the y-coordinates of A and B are the same (i.e., f(a) = f(b)), then there must be at least one point 'c' between a and b where the tangent to the curve is horizontal. A horizontal tangent has a slope of zero, which corresponds to f'(c) = 0.
4. How do you verify if Rolle's Theorem applies to a given function and interval?
To verify Rolle's Theorem for a function f(x) on an interval [a, b], you must follow these steps:
- Step 1: Check for Continuity. Confirm that the function is continuous on the closed interval [a, b]. For polynomials, trigonometric, and exponential functions, this is generally true within their domains.
- Step 2: Check for Differentiability. Find the derivative f'(x) and ensure it exists for all x in the open interval (a, b).
- Step 3: Check Endpoint Values. Calculate f(a) and f(b) to verify that f(a) = f(b).
- Step 4: Find 'c'. If all three conditions are satisfied, set f'(x) = 0 and solve for x. Any solution that falls within the open interval (a, b) is the value 'c' guaranteed by the theorem.
5. How is Rolle's Theorem related to Lagrange's Mean Value Theorem (LMVT)?
Rolle's Theorem is a special case of Lagrange's Mean Value Theorem (LMVT). LMVT states that for a continuous and differentiable function on [a, b], there is a point 'c' in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a). When we add the third condition from Rolle's Theorem, which is f(a) = f(b), the numerator f(b) - f(a) becomes zero. This simplifies the LMVT formula to f'(c) = 0, which is the exact conclusion of Rolle's Theorem.
6. What is the importance of Rolle's Theorem in finding the roots of a derivative?
A key application of Rolle's Theorem is in locating the roots of a derivative. The theorem guarantees that if a function f(x) has two different roots (i.e., it crosses the x-axis at two points, say x=a and x=b), then its derivative, f'(x), must have at least one root between 'a' and 'b'. This is because f(a) = 0 and f(b) = 0 satisfies the condition f(a) = f(b), implying there must be a point 'c' between them where the slope is zero (f'(c) = 0).
7. Why is Rolle's Theorem not applicable for the function f(x) = |x| on the interval [-1, 1]?
Rolle's Theorem is not applicable to f(x) = |x| (the absolute value function) on [-1, 1] because it fails one of the essential conditions. While the function is continuous on [-1, 1] and f(-1) = 1 = f(1), it is not differentiable at x = 0. Since x = 0 lies within the open interval (-1, 1), the condition of differentiability on the entire open interval is not met, making the theorem invalid for this case.
8. If the derivative of a function is zero at a point, does it automatically mean Rolle's Theorem must apply?
No, the converse of Rolle's Theorem is not necessarily true. A function can have a point 'c' where its derivative f'(c) = 0, but the theorem's conditions may not be met on a given interval [a, b]. For example, the function f(x) = x³ has a derivative f'(x) = 3x², and f'(0) = 0. However, on the interval [-1, 2], we have f(-1) = -1 and f(2) = 8, so f(-1) ≠ f(2). Even though a point with a zero derivative exists, Rolle's Theorem itself does not apply to this interval because the endpoint values are not equal.
