Check sibling questions

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Transcript

Ex 5.8, 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.