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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 1 Verify Rolle’s theorem for the function 𝑓 (π‘₯) = π‘₯2 + 2π‘₯ – 8, π‘₯ ∈ [4, 2].Let’s check conditions of Rolle’s theorem Condition 1 We need to check if 𝑓(π‘₯) is continuous at [–4, 2] Since 𝒇(𝒙)=π‘₯2 + 2π‘₯ – 8 is a polynomial & Every polynomial function is continuous for all π‘₯ βˆˆπ‘… ∴ 𝑓(π‘₯)is continuous at π‘₯∈[–4, 2] Conditions of Rolle’s theorem 𝑓(π‘₯) is continuous at [π‘Ž , 𝑏] 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) 𝑓(π‘Ž) = 𝑓(𝑏) If all 3 conditions are satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Condition 2 We need to check if 𝑓(π‘₯) is differentiable at (βˆ’4 , 2) Since 𝒇(𝒙) =π‘₯2 + 2π‘₯ – 8 is a polynomial . & Every polynomial function is differentiable for all π‘₯ βˆˆπ‘… ∴ 𝑓(π‘₯) is differentiable at (βˆ’4 , 2) Condition 3 We need to check if 𝒇(𝒂) = 𝒇(𝒃), for a = βˆ’4, b = 2 𝒇(βˆ’πŸ’) 𝒇(βˆ’πŸ’) = (βˆ’4)^2+2(βˆ’4)βˆ’8 = 16 βˆ’ 8 βˆ’ 8 = 0 𝒇(𝟐) 𝒇(𝟐)" = " (2)^2+2(2)βˆ’8" " "= " 4+4βˆ’8" = 0" Hence, 𝑓(βˆ’4) = 𝑓(2) Now, 𝑓(π‘₯) = π‘₯2 + 2π‘₯ – 8" " 𝒇^β€² (𝒙) = 2π‘₯+2βˆ’0 𝑓^β€² (π‘₯) = 2π‘₯+2 𝒇^β€² (𝒄) = πŸπ’„+𝟐 Since all three conditions satisfied 𝒇^β€² (𝒄) = 𝟎 2𝑐+2 = 0 2c = – 2 c = (βˆ’2)/2 c = βˆ’1 Since c = βˆ’1 ∈(βˆ’4 , 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.