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Ex 5.8

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Last updated at March 16, 2023 by Teachoo

Ex 5.8, 5 Verify Mean Value Theorem, if π (π₯) = π₯3 β 5π₯2 β 3π₯ in the interval [a, b], where a = 1 and b = 3. Find all π β (1, 3) for which π β²(π) = 0.π (π₯) = π₯3 β 5π₯2 β 3π₯ in [a, b], where a = 1 and b = 3 Condition 1 π (π₯) = π₯3 β 5π₯2 β 3π₯ π(π₯) is a polynomial & every polynomial function is continuous β΄ π(π₯) is continuous at π₯β[1, 3] Conditions of Mean value theorem π(π₯) is continuous at (π, π) π(π₯) is derivable at (π, π) If both conditions satisfied, then there exist some c in (π, π) such that πβ²(π) = (π(π) β π(π))/(π β π) Condition 2 π(π₯) = π₯3 β 5π₯2 β 3π₯ π(π₯) is a polynomial & every polynomial function is differentiable β΄ π(π₯) is differentiable at π₯β[1, 3] Now, π(π₯)" = " π₯3 β 5π₯2 β 3π₯ π^β² (π₯)" = 3" π₯2 β10π₯ β 3 π₯β[1, 3] So, πβ²(π) = " 3" π^2β10πβ3 Also, π(π₯)" = " π₯3 β 5π₯2 β 3π₯ π(π)" = " π(1) = (1)^3β5(1)^2β3(1) = 1β5β3 = β7 π(π)" = " π(3) = (3)^3β5(3)^2β3(3) = 27β45β9 = β27 By Mean Value Theorem π^β² (π) = (π(π) β π(π))/(π β π) "3" π^2β10πβ3 = (β27 β (β7))/(3 β 1) "3" π^2β10πβ3 = (β27 + 7)/2 "3" π^2β10πβ3 = (β20)/2 "3" π^2β10πβ3 = β10 "3" π^2β10πβ3+10 = 0 "3" π^2β10π+7 = 0 "3" π^2β3πβ7π+7 = 0 "3" π(πβ1)β7(πβ1) = 0 (3πβ7)(πβ1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = π/π β[1, 3] Thus, Mean Value Theorem is verified. From our question Find all π β (1, 3) for which π β²(π) = 0. We need to find cβ[1, 3] For which π^β² (π) = 0 π^β² (π) = 0 "3" π^2β10πβ3 = 0 The above equation is of the form π΄π₯^2+π΅π₯+πΆ x = (βπ΅ Β± β(π΅^2 β4π΄πΆ) )/2π΄ c = (β(β10) Β± β((β10)^2 β 4(β3)(3) ) )/2π΄ c = (10 Β± β(100 + 36) )/2(β3) c = (10 Β± β136 )/6 c = (10 Β± β(2 Γ 2 Γ 34))/6 c = (10 Β± 2β34)/6 c = 2(5 Β± β34 )/6 c = (5 Β± β34)/3 So, c = (5 + β34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 β β34)/3 c = (5 β 5.83)/3 c = (β0.83)/3 c = β0.28 Thus, c = 3.61 & c = β0.28 But both values do not lie between [1, 3] Hence, there exists no value of πβ[1, 3] for which π^β² (π) = 0