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Ex 6.5
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii) You are here
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5, 26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at March 16, 2023 by Teachoo
Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (viii) f(π₯) = π₯β(1βπ₯), π₯ > 0Finding fβ(π) fβ(π₯)=π(π₯β(1 β π₯))/ππ₯ fβ(π₯)=π(π₯)/ππ₯ . β(1βπ₯) + π(β(1 β π₯))/ππ₯ . π₯ = 1 . β(1βπ₯) + 1/(2β(1 β π₯)) . π(1 β π₯)/ππ₯ . π₯ = β(1βπ₯) + 1/(2β(1 β π₯)) (0 β1) . π₯ = β(1βπ₯) β π₯/(2β(1 β π₯)) = (2(β(1 β π₯) )^2β π₯)/(2β(1 β π₯)) = (2(1 β π₯) β π₯)/(2β(1 β π₯)) = (2 β 2π₯ β π₯)/(2β(1 β π₯)) = (2 β 3π₯)/(2β(1 β π₯)) Putting fβ(π)=π (2 β 3π₯)/(2β(1 β π₯))=0 2 β 3π₯ = 0 Γ 2β(1βπ₯) 2 β 3π₯=0 β 3π₯=β2 π₯ =2/3 Finding fββ(π) fβ(π₯)=(2 β 3π₯)/(2β(1 β π₯)) fββ(π₯)=π/ππ₯ ((2 β 3π₯)/(2β(1 β π₯))) = 1/2 [(π(2 β 3π₯)/ππ₯ . β(1 β π₯) β π(β(1 β π₯))/ππ₯ . (2 β 3π₯))/(β(1 β π₯))^2 ] = 1/2 [((0 β 3) β(1 β π₯) β 1/(2β(1 β π₯)) . π(1 β π₯)/ππ₯ . (2 β 3π₯))/((1 β π₯) )] Using quotient rule as (π’/π£)^β²=(π’^β² π£ β π£^β² π’)/π£^2 = 1/2 [(β3β(1 β π₯) β 1/(2β(1 β π₯)) (0 β 1) . (2 β 3π₯))/((1 β π₯) )] = 1/2 [(β3β(1 β π₯) + (2 β 3π₯)/(2β(1 β π₯)) )/(1 β π₯)] = 1/2 [((β3β(1 β π₯)) (2β(1 β π₯)) + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/2 [(β6(1 β π₯) + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/2 [(β6 + 6π₯ + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/4 [(β4 + 3π₯ )/(1 + π₯)^(3/2) ] Hence, fββ(π₯)=1/4 [(β4 + 3π₯ )/(1 + π₯)^(3/2) ] Putting π₯=2/3 fββ(2/3)=1/4 [(β4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(β4 + 2)/(5/3)^(3/2) ] =1/4 [(β2)/(5/3)^(3/2) ] = (β 1)/2 (3/5)^(3/2) < 0 Since fββ(π₯)<0 when π₯ = 2/3 Hence, π₯=2/3 is the maxima Finding Maximum value of f(π)=πβ(πβπ) Putting π₯=2/3 f(2/3)=2/3 β(1β2/3) =2/3 β((3 β 2)/3) Finding Maximum value of f(π)=πβ(πβπ) Putting π₯=2/3 f(π₯) = π₯β(1βπ₯) f(2/3)=2/3 β(1β2/3) =2/3 β((3 β 2)/3) =2/3 β(1/3) =2/(3β3) =2/(3β3) Γ β3/β3 =(2β3)/9 Maximum value of f(π) is (πβπ)/π at x = π/π