Rolle's and Mean Value Theorem

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Question 2 Examine if Rolleβs theorem is applicable to the functions. Can you say some thing about the converse of Rolleβs theorem from this function? (πππ) π (π₯) = π₯2 β 1 πππ π₯ β [1, 2]π (π₯) = π₯2 β 1 πππ π₯ β [1 , 2] Condition 1 π(π₯) = π₯2 β 1 π(π₯) is a polynomial & Every polynomial function is continuous β π(π₯) is continuous at π₯β[1, 2] Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 π(π₯)=π₯2 β 1 π(π₯) is a polynomial & Every polynomial function is differentiable β π(π₯) is differentiable at π₯β[1, 2] Condition 3 π(π₯) = π₯2 β 1 π(1) = (1)^2+(1) = 1 β 1 = 0 & π(2) = (2)^2β1= 4β1 = 3 Since π(π) β  π(π) Thus third condition of Rolleβs Theorem is not satisfied. Therefore Rolleβs theorem is not applicable Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Converse of Rolleβs Theorem If π [π, π]βπ for some πβ[π, π] for which π^β² (π)=0 then (i) π(π) = π(π) (ii) π is continuous at [π, π] (iii) & Differentiable at [π, π] Now, π(π₯)=π₯^2β1 π^β² (π₯)=2π₯ π^β² (π)=2π Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 If π^β² (π)=0 2π=0 π=0 Since π=0 does not belong in (1, 2) i.e. c = 0 β (1 , 2) β There is no value of c for which π^β² (π)=0 β΄ Converse of Rolleβs Theorem is also not applicable.

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.