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

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.8, 1 Verify Rolleโ€™s theorem for the function ๐‘“ (๐‘ฅ) = ๐‘ฅ2 + 2๐‘ฅ โ€“ 8, ๐‘ฅ โˆˆ [โ€“ 4, 2]. ๐‘“ (๐‘ฅ) = ๐‘ฅ2 + 2๐‘ฅ โ€“ 8, ๐‘ฅ โˆˆ [โ€“ 4, 2]. Rolleโ€™s theorem is satisfied if Condition 1 ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 + 2๐‘ฅ โ€“ 8 is continuous at โˆ’4 , 2๏ทฏ Since ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 + 2๐‘ฅ โ€“ 8 is a polynomial & Every polynomial function is continuous for all ๐‘ฅ โˆˆ๐‘… โ‡’ ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 + 2๐‘ฅ โ€“ 8 is continuous at ๐‘ฅโˆˆ[โ€“ 4, 2] Condition 2 ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 + 2๐‘ฅ โ€“ 8 is differentiable at โˆ’4 , 2๏ทฏ ๐‘“(๐‘ฅ) =๐‘ฅ2 + 2๐‘ฅ โ€“ 8 is a polynomial . & Every polynomial function is differentiable for all ๐‘ฅ โˆˆ๐‘… therefore ๐‘“(๐‘ฅ) is differentiable at โˆ’4 , 2๏ทฏ Condition 3 ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 + 2๐‘ฅ โ€“ 8 ๐‘“ โˆ’4๏ทฏ = โˆ’4๏ทฏ๏ทฎ2๏ทฏ+2 โˆ’4๏ทฏโˆ’8 = 16 โˆ’ 8 โˆ’ 8 = 16 โˆ’ 16 = 0 & ๐‘“(2) = 2๏ทฏ๏ทฎ2๏ทฏ+2 2๏ทฏโˆ’8 = 4+4โˆ’8 = 8โˆ’8 = 0 Hence ๐‘“ โˆ’4๏ทฏ = ๐‘“ 2๏ทฏ Now, ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ2 + 2๐‘ฅ โ€“ 8 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘ฅ๏ทฏ = 2๐‘ฅ+2โˆ’0 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘ฅ๏ทฏ = 2๐‘ฅ+2 ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘ฅ๏ทฏ = 2๐‘+2 Since all three condition satisfied ๐‘“๏ทฎโ€ฒ๏ทฏ ๐‘๏ทฏ = 0 2๐‘+2 = 0 2c = โ€“ 2 ๐‘ = โˆ’ 2๏ทฎ2๏ทฏ = โˆ’1 Value of c = โˆ’1 โˆˆ โˆ’4 , 2๏ทฏ Thus, Rolleโ€™s Theorem is satisfied.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.