Ex 6.5

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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(1βπ₯), π₯ > 0Finding fβ(π) fβ(π₯)=π(π₯β(1 β π₯))/ππ₯ fβ(π₯)=π(π₯)/ππ₯ . β(1βπ₯) + π(β(1 β π₯))/ππ₯ . π₯ = 1 . β(1βπ₯) + 1/(2β(1 β π₯)) . π(1 β π₯)/ππ₯ . π₯ = β(1βπ₯) + 1/(2β(1 β π₯)) (0 β1) . π₯ = β(1βπ₯) 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(1βπ₯) 2 β 3π₯=0 β 3π₯=β2 π₯ =2/3 Finding fββ(π) fβ(π₯)=(2 β 3π₯)/(2β(1 β π₯)) fββ(π₯)=π/ππ₯ ((2 β 3π₯)/(2β(1 β π₯))) = 1/2 [(π(2 β 3π₯)/ππ₯ . β(1 β π₯) β π(β(1 β π₯))/ππ₯ . (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₯)/ππ₯ . (2 β 3π₯))/((1 β π₯) )] Using quotient rule as (π’/π£)^β²=(π’^β² π£ β π£^β² π’)/π£^2 = 1/2 [(β3β(1 β π₯) β 1/(2β(1 β π₯)) (0 β 1) . (2 β 3π₯))/((1 β π₯) )] = 1/2 [(β3β(1 β π₯) + (2 β 3π₯)/(2β(1 β π₯)) )/(1 β π₯)] = 1/2 [((β3β(1 β π₯)) (2β(1 β π₯)) + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/2 [(β6(1 β π₯) + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/2 [(β6 + 6π₯ + 2 β 3π₯ )/(2(1 β π₯) β(1 β π₯))] = 1/4 [(β4 + 3π₯ )/(1 + π₯)^(3/2) ] Hence, fββ(π₯)=1/4 [(β4 + 3π₯ )/(1 + π₯)^(3/2) ] Putting π₯=2/3 fββ(2/3)=1/4 [(β4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(β4 + 2)/(5/3)^(3/2) ] =1/4 [(β2)/(5/3)^(3/2) ] = (β 1)/2 (3/5)^(3/2) < 0 Since fββ(π₯)<0 when π₯ = 2/3 Hence, π₯=2/3 is the maxima Finding Maximum value of f(π)=πβ(πβπ) Putting π₯=2/3 f(2/3)=2/3 β(1β2/3) =2/3 β((3 β 2)/3) Finding Maximum value of f(π)=πβ(πβπ) Putting π₯=2/3 f(π₯) = π₯β(1βπ₯) f(2/3)=2/3 β(1β2/3) =2/3 β((3 β 2)/3) =2/3 β(1/3) =2/(3β3) =2/(3β3) Γ β3/β3 =(2β3)/9 Maximum value of f(π) is (πβπ)/π at x = π/π