Example 42 - Verify Rolle’s theorem for y = x2 + 2, a = -2

Example 42 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 42 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Example 42 Verify Rolle’s theorem for the function y = x2 + 2, a = – 2 and b = 2. y = x2 + 2, a = –2 and b = 2 Let 𝑓(π‘₯) = π‘₯^2+2 Rolle’s theorem is satisfied if Condition 1 Since 𝑓(π‘₯) is a polynomial, it is continuous ∴ 𝑓(π‘₯) is continuous at (βˆ’2 , 2) Conditions of Rolle’s theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is differentiable at (π‘Ž , 𝑏) 𝑓(π‘Ž) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Condition 2 Since 𝑓(π‘₯) is a polynomial, it is differentiable ∴ 𝑓(π‘₯) is differentiable at (βˆ’2 , 2) Also, 𝒇(βˆ’πŸ) = (βˆ’2)^2+2 = 4+2 = 6 𝒇(𝟐) = 2^2+2 = 4+2 = 6 Hence, 𝒇(𝟐) = 𝒇(βˆ’πŸ) Conditions of Rolle’s theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is differentiable at (π‘Ž , 𝑏) 𝑓(π‘Ž) = 𝑓(𝑏) If all 3 conditions satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Also, 𝑓(π‘₯) = π‘₯^2+2 𝑓^β€² (π‘₯) = 2x So, 𝒇^β€² (𝒄) = πŸπ’„ Since all 3 conditions are satisfied 𝒇^β€² (𝒄) = 𝟎 2𝑐 = 0 𝒄 = 𝟎 Value of c i.e. 0 lies between βˆ’2 and 2. Hence c = 0 ∈ (βˆ’πŸ, 𝟐) Thus, Rolle’s theorem is satisfied.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.