Ex 5.8, 4 - Verify Mean Value Theorem f(x) = x2 - 4x - 3

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

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

Transcript

Ex 5.8, 4 Verify Mean Value Theorem, if ๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 in the interval [๐‘Ž, ๐‘], where ๐‘Ž= 1 ๐‘Ž๐‘›๐‘‘ ๐‘= 4 ๐‘“ (๐‘ฅ) = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘ฅโˆˆ[๐‘Ž, ๐‘] where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 ๐‘“(๐‘ฅ) is continuous ๐‘“(๐‘ฅ)=๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“(๐‘ฅ) is a polynomial & Every polynomial function is continuous โ‡’ ๐‘“(๐‘ฅ) is continuous at ๐‘ฅโˆˆ[1, 4] Conditions of Mean value theorem ๐‘“(๐‘ฅ) is continuous at (๐‘Ž , ๐‘) ๐‘“(๐‘ฅ) is derivable at (๐‘Ž , ๐‘) If both conditions satisfied, then there exist some c in (๐‘Ž , ๐‘) such that ๐‘“โ€ฒ(๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) Conditions of Mean value theorem ๐‘“(๐‘ฅ) is continuous at (๐‘Ž , ๐‘) ๐‘“(๐‘ฅ) is derivable at (๐‘Ž , ๐‘) If both conditions satisfied, then there exist some c in (๐‘Ž , ๐‘) such that ๐‘“โ€ฒ(๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) Condition 2 If ๐‘“(๐‘ฅ) is differentiable ๐‘“(๐‘ฅ) = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“(๐‘ฅ) is a polynomial & Every polynomial function is differentiable โ‡’ ๐‘“(๐‘ฅ) is differentiable at ๐‘ฅโˆˆ[1, 4] Condition 3 ๐‘“(๐‘ฅ) = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“^โ€ฒ (๐‘ฅ) = 2๐‘ฅโˆ’4 ๐‘“^โ€ฒ (๐‘) = 2๐‘โˆ’4 Conditions of Mean value theorem ๐‘“(๐‘ฅ) is continuous at (๐‘Ž , ๐‘) ๐‘“(๐‘ฅ) is derivable at (๐‘Ž , ๐‘) If both conditions satisfied, then there exist some c in (๐‘Ž , ๐‘) such that ๐‘“โ€ฒ(๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) ๐‘“(๐‘Ž) = ๐‘“(1) = (1)^2โˆ’4(1)โˆ’3 = 1 โˆ’ 4 โˆ’ 3 = โˆ’6 ๐‘“(๐‘) = ๐‘“(4) = (4)^2โˆ’4(4)โˆ’3 = 16 โˆ’ 16 โˆ’ 3 = โˆ’ 3 By Mean Value Theorem ๐‘“^โ€ฒ (๐‘) = (๐‘“(๐‘) โˆ’ ๐‘“(๐‘Ž))/(๐‘ โˆ’ ๐‘Ž) ๐‘“^โ€ฒ (๐‘) = (โˆ’3 โˆ’ (โˆ’6))/(4 โˆ’ 1) ๐‘“^โ€ฒ (๐‘) = (โˆ’3 + 6)/3 ๐‘“^โ€ฒ (๐‘) = 3/3 ๐‘“^โ€ฒ (๐‘) = 1 2c โˆ’ 4 = 1 2c = 1 + 4 2c = 5 c = 5/2 Value of c = 5/2 which is lies between (1, 4) c = ๐Ÿ“/๐Ÿโˆˆ(๐Ÿ, ๐Ÿ’) Hence Mean Value Theorem 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.