(i) Rolle's theorem states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), with f(a) = f(b), then there exists at least one value c in the interval (a,b) where the derivative f'(c) = 0. (ii) The example function f(x)=x^2-4x+4 on the interval [1,3] satisfies the conditions of Rolle's theorem by being continuous, differentiable, and having equal values at the endpoints. The value of c where f'(c)=0 is found to be 2. (iii

1. BRAINWARE UNIVERSITY
❖Program name : B.Tech InRobotics & Automation
❖Course code : BSCR102
❖Course Name :Calculus and Linear Algebra
❖Semester/Year : 1st
❖Name Of the Student : AMIT MANNA
❖Student Code : BWU/BEC/23/008
❖Name of The Department : Mathematics
Presentationno - 01

2. TOPIC : Mean Value
Theorem
WELCOME to my
presentation

3. introduction
Rolles theoremis a fundamental theorem in calculus, named after Scottish
mathematician Jhon Rolls. The theorem is an important result related to the
behaviourof a differentiable function on a closed interval. Here's a statement of
Rolls Theorem:
ROLLES THEOREM :
Let, f be a real value function, where I = [a,b] & f satisfying the
following condition -
(i) f is continuous in the close interval [a,b].
(ii) f is differentiable in the open interval (a,b) and f(x) is exists for
x belongs to (a,b).
(iii) f(a) = f(b), Then there exists at least one value c, a< c< b, such
that f’(c) = 0 .

4. Geometrical interpretation
of rolle’s theorem.
A function f which is continuous on the closed interval [a , b].
Can be represented graphically by a curve without break.
Since f is differentiable on (a , b), the curve has a tangent at each
point on the arc AB where A = {a, f(a)} B = {b, f(b)}.
Also, since f(a) = f(b), so the ordinates of the two points A and B
are same.

5. EXAMPLE.
Q. Verify Rolle's Theorem For The function f(x)=x2−4x+4
on the interval [1,3]. And find the value of c.
Ans,
 CONTINUITY : The function f(x) is a polynomial, and polynomials are continuous
everywhere. Therefore, f(x) is continuous on [1,3].
 DIFFERENTIABILITY:The derivative of f(x) is f′(x)=2x−4f′(x)=2x−4, which is defined
everywhere. So, f(x) is differentiableon (1,3).
 EQUAL VALUE AT END POINTS : Evaluatef(1) and f(3):
 f(1)=12−4⋅1+4=1−4+4=1f(1)=12−4⋅1+4=1−4+4=1
 f(3)=32−4⋅3+4=9−12+4=1f(3)=32−4⋅3+4=9−12+4=1
Since f(1)=f(3), the conditions of Rolle's Theorem are satisfied.
Now, we find the derivativeof f(x): f′(x)=2x−4
Setting f′(x) equal to zero and solving for x:
2x−4=0
Or, 2x=4
Or, x=2
So, according to Rolle's Theorem, there exists at least one c in the open interval
(1,3) such that f′(c)=0. In this case, c= 2.
Therefore f(x) satisfies all the conditionsof Rolle’s theorem.