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Ex 5.8, 5 - Verify Mean Value Theorem f(x) = x3 - 5x2 - 3x

Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 6 Ex 5.8, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

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Question 5 Verify Mean Value Theorem, if š‘“ (š‘„) = š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ in the interval [a, b], where a = 1 and b = 3. Find all š‘ āˆˆ (1, 3) for which š‘“ ā€²(š‘) = 0.š‘“ (š‘„) = š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ in [a, b], where a = 1 and b = 3 Condition 1 š‘“ (š‘„) = š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ š‘“(š‘„) is a polynomial & every polynomial function is continuous āˆ“ š‘“(š‘„) is continuous at š‘„āˆˆ[1, 3] Conditions of Mean value theorem š‘“(š‘„) is continuous at (š‘Ž, š‘) š‘“(š‘„) is derivable at (š‘Ž, š‘) If both conditions satisfied, then there exist some c in (š‘Ž, š‘) such that š‘“ā€²(š‘) = (š‘“(š‘) āˆ’ š‘“(š‘Ž))/(š‘ āˆ’ š‘Ž) Condition 2 š‘“(š‘„) = š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ š‘“(š‘„) is a polynomial & every polynomial function is differentiable āˆ“ š‘“(š‘„) is differentiable at š‘„āˆˆ[1, 3] Now, š‘“(š‘„)" = " š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ š‘“^ā€² (š‘„)" = 3" š‘„2 ā€“10š‘„ ā€“ 3 š‘„āˆˆ[1, 3] So, š‘“ā€²(š‘) = " 3" š‘^2āˆ’10š‘āˆ’3 Also, š‘“(š‘„)" = " š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ š‘“(š‘Ž)" = " š‘“(1) = (1)^3āˆ’5(1)^2āˆ’3(1) = 1āˆ’5āˆ’3 = āˆ’7 š‘“(š‘)" = " š‘“(3) = (3)^3āˆ’5(3)^2āˆ’3(3) = 27āˆ’45āˆ’9 = āˆ’27 By Mean Value Theorem š‘“^ā€² (š‘) = (š‘“(š‘) āˆ’ š‘“(š‘Ž))/(š‘ āˆ’ š‘Ž) "3" š‘^2āˆ’10š‘āˆ’3 = (āˆ’27 āˆ’ (āˆ’7))/(3 āˆ’ 1) "3" š‘^2āˆ’10š‘āˆ’3 = (āˆ’27 + 7)/2 "3" š‘^2āˆ’10š‘āˆ’3 = (āˆ’20)/2 "3" š‘^2āˆ’10š‘āˆ’3 = āˆ’10 "3" š‘^2āˆ’10š‘āˆ’3+10 = 0 "3" š‘^2āˆ’10š‘+7 = 0 "3" š‘^2āˆ’3š‘āˆ’7š‘+7 = 0 "3" š‘(š‘āˆ’1)āˆ’7(š‘āˆ’1) = 0 (3š‘āˆ’7)(š‘āˆ’1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = šŸ•/šŸ‘ āˆˆ[1, 3] Thus, Mean Value Theorem is verified. From our question Find all š‘ āˆˆ (1, 3) for which š‘“ ā€²(š‘) = 0. We need to find cāˆˆ[1, 3] For which š‘“^ā€² (š‘) = 0 š‘“^ā€² (š‘) = 0 "3" š‘^2āˆ’10š‘āˆ’3 = 0 The above equation is of the form š“š‘„^2+šµš‘„+š¶ x = (āˆ’šµ Ā± āˆš(šµ^2 āˆ’4š“š¶) )/2š“ c = (āˆ’(āˆ’10) Ā± āˆš((āˆ’10)^2 āˆ’ 4(āˆ’3)(3) ) )/2š“ c = (10 Ā± āˆš(100 + 36) )/2(āˆ’3) c = (10 Ā± āˆš136 )/6 c = (10 Ā± āˆš(2 Ɨ 2 Ɨ 34))/6 c = (10 Ā± 2āˆš34)/6 c = 2(5 Ā± āˆš34 )/6 c = (5 Ā± āˆš34)/3 So, c = (5 + āˆš34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 āˆ’ āˆš34)/3 c = (5 āˆ’ 5.83)/3 c = (āˆ’0.83)/3 c = āˆ’0.28 Thus, c = 3.61 & c = ā€“0.28 But both values do not lie between [1, 3] Hence, there exists no value of šœāˆˆ[1, 3] for which š‘“^ā€² (š‘) = 0

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.