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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


Transcript

Ex 5.8, 4 Verify Mean Value Theorem, if 𝑓 (π‘₯) = π‘₯2 – 4π‘₯ – 3 in the interval [π‘Ž, 𝑏], where π‘Ž= 1 π‘Žπ‘›π‘‘ 𝑏= 4 𝑓 (π‘₯) = π‘₯2 – 4π‘₯ – 3 π‘₯∈[π‘Ž, 𝑏] where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 𝑓(π‘₯) is continuous 𝑓(π‘₯)=π‘₯2 – 4π‘₯ – 3 𝑓(π‘₯) is a polynomial & Every polynomial function is continuous β‡’ 𝑓(π‘₯) is continuous at π‘₯∈[1, 4] Conditions of Mean value theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) If both conditions satisfied, then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) Conditions of Mean value theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) If both conditions satisfied, then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) Condition 2 If 𝑓(π‘₯) is differentiable 𝑓(π‘₯) = π‘₯2 – 4π‘₯ – 3 𝑓(π‘₯) is a polynomial & Every polynomial function is differentiable β‡’ 𝑓(π‘₯) is differentiable at π‘₯∈[1, 4] Condition 3 𝑓(π‘₯) = π‘₯2 – 4π‘₯ – 3 𝑓^β€² (π‘₯) = 2π‘₯βˆ’4 𝑓^β€² (𝑐) = 2π‘βˆ’4 Conditions of Mean value theorem 𝑓(π‘₯) is continuous at (π‘Ž , 𝑏) 𝑓(π‘₯) is derivable at (π‘Ž , 𝑏) If both conditions satisfied, then there exist some c in (π‘Ž , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) 𝑓(π‘Ž) = 𝑓(1) = (1)^2βˆ’4(1)βˆ’3 = 1 βˆ’ 4 βˆ’ 3 = βˆ’6 𝑓(𝑏) = 𝑓(4) = (4)^2βˆ’4(4)βˆ’3 = 16 βˆ’ 16 βˆ’ 3 = βˆ’ 3 By Mean Value Theorem 𝑓^β€² (𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) 𝑓^β€² (𝑐) = (βˆ’3 βˆ’ (βˆ’6))/(4 βˆ’ 1) 𝑓^β€² (𝑐) = (βˆ’3 + 6)/3 𝑓^β€² (𝑐) = 3/3 𝑓^β€² (𝑐) = 1 2c βˆ’ 4 = 1 2c = 1 + 4 2c = 5 c = 5/2 Value of c = 5/2 which is lies between (1, 4) c = πŸ“/𝟐∈(𝟏, πŸ’) Hence Mean Value Theorem satisfied

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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.