Ex 5.8, 1 - Verify Rolle’s theorem for f(x) = x2 + 2x - 8

Ex 5.8, 1 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.8, 1 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.8, 1 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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

Transcript

Ex 5.8, 1 Verify Rolle’s theorem for the function 𝑓 (π‘₯) = π‘₯2 + 2π‘₯ – 8, π‘₯ ∈ [– 4, 2]. 𝑓 (π‘₯) = π‘₯2 + 2π‘₯ – 8, π‘₯ ∈ [– 4, 2]. Rolle’s theorem is satisfied if Condition 1 𝑓(π‘₯)=π‘₯2 + 2π‘₯ – 8 is continuous at (βˆ’4 , 2) Since 𝑓(π‘₯)=π‘₯2 + 2π‘₯ – 8 is a polynomial & Every polynomial function is continuous for all π‘₯ βˆˆπ‘… β‡’ 𝑓(π‘₯)=π‘₯2 + 2π‘₯ – 8 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 satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Condition 2 𝑓(π‘₯)=π‘₯2 + 2π‘₯ – 8 is differentiable at (βˆ’4 , 2) 𝑓(π‘₯) =π‘₯2 + 2π‘₯ – 8 is a polynomial . & Every polynomial function is differentiable for all π‘₯ βˆˆπ‘… therefore 𝑓(π‘₯) is differentiable at (βˆ’4 , 2) Condition 3 𝑓(π‘₯) = π‘₯2 + 2π‘₯ – 8" " 𝑓(βˆ’4) = (βˆ’4)^2+2(βˆ’4)βˆ’8 = 16 βˆ’ 8 βˆ’ 8 = 16 βˆ’ 16 = 0 & 𝑓(2) = (2)^2+2(2)βˆ’8 = 4+4βˆ’8 = 8βˆ’8 = 0 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 satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 Hence 𝑓(βˆ’4) = 𝑓(2) Now, 𝑓(π‘₯) = π‘₯2 + 2π‘₯ – 8" " 𝑓^β€² (π‘₯) = 2π‘₯+2βˆ’0 𝑓^β€² (π‘₯) = 2π‘₯+2 𝑓^β€² (π‘₯) = 2𝑐+2 Since all three condition satisfied 𝑓^β€² (𝑐) = 0 2𝑐+2 = 0 2c = – 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 satisfied then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = 0 𝑐 = (βˆ’ 2)/2 = βˆ’1 Value of 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 10 years. He provides courses for Maths and Science at Teachoo.