Ex 5.8, 4 - Ex 5.8, 4
Verify Mean Value Theorem, if ๐‘“ (๐‘ฅ) = ๐‘ฅ2 – 4๐‘ฅ – 3 in the interval [๐‘Ž, ๐‘], where ๐‘Ž= - Ex 5.8

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  1. Chapter 5 Class 12 Continuity and Differentiability
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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‘๏ทฏ where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 ๐‘“ ๐‘ฅ๏ทฏ is continuous ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“ ๐‘ฅ๏ทฏ is a polynomial & Every polynomial function is continuous โ‡’ ๐‘“ ๐‘ฅ๏ทฏ is continuous at ๐‘ฅโˆˆ[1, 4] Condition 2 If ๐‘“ ๐‘ฅ๏ทฏ is differentiable ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“ ๐‘ฅ๏ทฏ is a polynomial & Every polynomial function is differentiable โ‡’ ๐‘“ ๐‘ฅ๏ทฏ is differentiable at ๐‘ฅโˆˆ 1, 4๏ทฏ Condition 3 ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 โ€“ 4๐‘ฅ โ€“ 3 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘ฅ๏ทฏ = 2๐‘ฅโˆ’4 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = 2๐‘โˆ’4 ๐‘“(๐‘Ž) = ๐‘“(1) = 1๏ทฏ๏ทฎ2๏ทฏโˆ’4 1๏ทฏโˆ’3 = 1 โˆ’ 4 โˆ’ 3 = โˆ’6 ๐‘“ ๐‘๏ทฏ = ๐‘“ 4๏ทฏ = 4๏ทฏ๏ทฎ2๏ทฏโˆ’4 4๏ทฏโˆ’3 = 16 โˆ’ 16 โˆ’ 3 = โˆ’ 3 By Mean Value Theorem ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = ๐‘“ ๐‘๏ทฏ โˆ’ ๐‘“ ๐‘Ž๏ทฏ๏ทฎ๐‘ โˆ’ ๐‘Ž๏ทฏ ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = โˆ’3 โˆ’ โˆ’6๏ทฏ๏ทฎ4 โˆ’ 1๏ทฏ ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = โˆ’3 + 6๏ทฎ3๏ทฏ ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = 3๏ทฎ3๏ทฏ ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = 1 2c โˆ’ 4 = 1 2c = 1 + 4 2c = 5 c = 5๏ทฎ2๏ทฏ Value of c = 5๏ทฎ2๏ทฏ which is lies between (1, 4) c = ๐Ÿ“๏ทฎ๐Ÿ๏ทฏโˆˆ ๐Ÿ, ๐Ÿ’๏ทฏ Hence Mean Value Theorem satisfied

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