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Chapter– 5: Continuity and Differentiability

Mathematics - XII

Sub-Topic: Rolle’s Theorem and

Mean Value Theorem

Unit – III: Calculus

Outline:

• Rolle’sTheorem
• Geometrical meaning of Rolle’s Theorem.
• Mean Value Theorem
• Geometrical meaning of Mean Value theorem
• Example
• Assignment

a

b

c

f’(c)=0

f(a)=f(b)

Rolle’s Theorem

Let f be a function, satisfying the conditions:

1. continuous at every point of a closed interval [a,b],
2. differentiable at every point of the open interval (a, b) and
3. f(a) = f(b), then

there exists at least one point c in the open interval (a, b),

where f’(c) = o

Geometrical Meaning of Rolle’s Theorem

x=a

x=b

Geometrically Rolle’s theorem shows that if the given function satisfies the conditions of Rolle’s Theorem, then there exist at least one point in the domain of function where tangent is parallel to x-axis.

Mean Value Theorem

a

b

c

f(a)

f(b)

Let f be a function, satisfying the conditions

1. continuous at every point of a closed interval [a,b],
2. differentiable at every point of the open interval (a, b), then

there exists at least one point c in the open interval (a, b), such that

​

Geometrical Meaning of Mean Value Theorem

Geometrically Mean Value Theorem shows that if the given function satisfies the conditions of Mean Value Theorem, then there exist at least one point in the domain of function where tangent is parallel to the secant joining any two points on the curve

Example:

Verify mean value theorem for the function

f (x) = (x – 3) (x – 6) (x – 9) in [3, 5].

Solution:

1. is continuous in [3, 5], as product of polynomial functions is a polynomial, which is continuous.

2. f ′(x) = 3x² – 36x + 99 exists in (3, 5) and hence derivable in (3, 5).

Thus, conditions of mean value theorem are satisfied.

Given function, f (x) = (x – 3) (x – 6) (x – 9)

Hence, there exists at least one c ∈ (3, 5) such that

Solution(Cont.)

Assignment:-