Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Ex 5.8, 5 Verify Mean Value Theorem, if 𝑓 (π‘₯) = π‘₯3 – 5π‘₯2 – 3π‘₯ in the interval [a, b], where a = 1 and b = 3. Find all 𝑐 ∈ (1, 3) for which 𝑓 β€²(𝑐) = 0.𝑓 (π‘₯) = π‘₯3 – 5π‘₯2 – 3π‘₯ in [a, b], where a = 1 and b = 3 Condition 1 𝑓 (π‘₯) = π‘₯3 – 5π‘₯2 – 3π‘₯ 𝑓(π‘₯) is a polynomial & every polynomial function is continuous ∴ 𝑓(π‘₯) is continuous at π‘₯∈[1, 3] 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 𝑓(π‘₯) = π‘₯3 – 5π‘₯2 – 3π‘₯ 𝑓(π‘₯) is a polynomial & every polynomial function is differentiable ∴ 𝑓(π‘₯) is differentiable at π‘₯∈[1, 3] Now, 𝑓(π‘₯)" = " π‘₯3 – 5π‘₯2 – 3π‘₯ 𝑓^β€² (π‘₯)" = 3" π‘₯2 –10π‘₯ – 3 π‘₯∈[1, 3] So, 𝑓′(𝑐) = " 3" 𝑐^2βˆ’10π‘βˆ’3 Also, 𝑓(π‘₯)" = " π‘₯3 – 5π‘₯2 – 3π‘₯ 𝑓(π‘Ž)" = " 𝑓(1) = (1)^3βˆ’5(1)^2βˆ’3(1) = 1βˆ’5βˆ’3 = βˆ’7 𝑓(𝑏)" = " 𝑓(3) = (3)^3βˆ’5(3)^2βˆ’3(3) = 27βˆ’45βˆ’9 = βˆ’27 By Mean Value Theorem 𝑓^β€² (𝑐) = (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) "3" 𝑐^2βˆ’10π‘βˆ’3 = (βˆ’27 βˆ’ (βˆ’7))/(3 βˆ’ 1) "3" 𝑐^2βˆ’10π‘βˆ’3 = (βˆ’27 + 7)/2 "3" 𝑐^2βˆ’10π‘βˆ’3 = (βˆ’20)/2 "3" 𝑐^2βˆ’10π‘βˆ’3 = βˆ’10 "3" 𝑐^2βˆ’10π‘βˆ’3+10 = 0 "3" 𝑐^2βˆ’10𝑐+7 = 0 "3" 𝑐^2βˆ’3π‘βˆ’7𝑐+7 = 0 "3" 𝑐(π‘βˆ’1)βˆ’7(π‘βˆ’1) = 0 (3π‘βˆ’7)(π‘βˆ’1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = πŸ•/πŸ‘ ∈[1, 3] Thus, Mean Value Theorem is verified. From our question Find all 𝑐 ∈ (1, 3) for which 𝑓 β€²(𝑐) = 0. We need to find c∈[1, 3] For which 𝑓^β€² (𝑐) = 0 𝑓^β€² (𝑐) = 0 "3" 𝑐^2βˆ’10π‘βˆ’3 = 0 The above equation is of the form 𝐴π‘₯^2+𝐡π‘₯+𝐢 x = (βˆ’π΅ Β± √(𝐡^2 βˆ’4𝐴𝐢) )/2𝐴 c = (βˆ’(βˆ’10) Β± √((βˆ’10)^2 βˆ’ 4(βˆ’3)(3) ) )/2𝐴 c = (10 Β± √(100 + 36) )/2(βˆ’3) c = (10 Β± √136 )/6 c = (10 Β± √(2 Γ— 2 Γ— 34))/6 c = (10 Β± 2√34)/6 c = 2(5 Β± √34 )/6 c = (5 Β± √34)/3 So, c = (5 + √34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 βˆ’ √34)/3 c = (5 βˆ’ 5.83)/3 c = (βˆ’0.83)/3 c = βˆ’0.28 Thus, c = 3.61 & c = –0.28 But both values do not lie between [1, 3] Hence, there exists no value of 𝐜∈[1, 3] for which 𝑓^β€² (𝑐) = 0

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

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.