Class 12 Maths Continuity Differentiability Mean Value Theorem

Mean Value Theorem

There are two fundamental results in Calculus. We will discuss both of them and also learn the geometric interpretation of these theorems.

Rolle’s Theorem:

Let f be a real valued function defined on the closed interval [a, b] such that

(i) It is continuous on the closed interval [a, b].

(ii) It is differentiable on the open interval (a, b)

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

 Then there exists a real number c є (a, b) such that f′(c) = 0.

Class_12_Maths_Continuity_&_Differentiability 

                             

In each of the above graph, the slope becomes zero at least at one point. That is precisely the claim of the Rolle’s theorem as the slope of the tangent at any point on the graph of y = f (x) is nothing but the derivative of f(x) at that point.

Problem: Verify Rolle’s Theorem for the function f(x) = x2 + 2x – 8, x ∈ [-4, 2]

Solution:

The given function f(x) = x2 + 2x – 8 being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).

f(-4) = (-4)2 + 2 * (-4) – 8 = 16 – 8 – 8 = 0

f(2) = 22 + 2 * 2 – 8 = 4 + 4 – 8 = 0

So, f (−4) = f (2) = 0

=> The value of f (x) at −4 and 2 coincides.

Rolle’s Theorem states that there is a point c ∈ (−4, 2) such that f’(c) = 0

Now, f(x) = x2 + 2x – 8

f’(x) = 2x + 2

Now, f’(c) = 0

=> 2c + 2 = 0

=> c = -1, where c ∈ (−4, 2)

Hence, Rolle’s Theorem is verified for the given function.

Mean Value Theorem:

Let f be a real valued function defined on the closed interval [a, b] such that

(i) It is continuous on the closed interval [a, b].

(ii) It is differentiable on the open interval (a, b)

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

Then there exists a real number c є (a, b) such that f′(c) = {f(b) – f(a)}/(b - a)

  Class_12_Maths_Continuity_&_Differentiability_Mean_Value_Therorem

This theorem is an extension of Rolle’s theorem. From the graph, it is clear that {f(b) – f(a)}/(b - a) is the slope of the secant drawn between (a, f(a)) and (b, f(b)). The theorem stats that there is a point c in (a, b) such that the tangent at (c, f(c)) is parallel to the secant between (a, f(a)) and (b, f(b)).  

Problem: Verify Mean Value Theorem, if f(x) = x2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Solution: The given function is f(x) = x2 – 4x – 3

The polynomial function f is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

f(1) = 12 – 4 * 1 – 3 = 1 – 4 – 3 = -6

f(4) = 42 – 4 * 4 – 3 = 16 – 16 – 3 = -3

So, {f(b) –f(a)}/(b - a) = {f(4) –f(1)}/(4 - 1) = {-3 – (-6)}/3 = (-3 + 6)/3 = 3/3 = 1

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f’(c) = 1

=> 2c – 4 = 1

=> 2c = 5

=> c = 5/2, where c = 5/2 ∈ (1, 4)

Hence, Mean Value Theorem is verified for the given function.

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