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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.