Ex 6.3, 3 (viii) - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 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(𝑥) = 𝑥√(1−𝑥), 𝑥 > 0= √(1−𝑥) + 1/(2√(1 − 𝑥)) (0 −1) . 𝑥 = √(1−𝑥) – 𝑥/(2√(1 − 𝑥)) = (2(√(1 − 𝑥) )^2− 𝑥)/(2√(1 − 𝑥)) = (2(1 − 𝑥) − 𝑥)/(2√(1 − 𝑥)) = (2 − 2𝑥 − 𝑥)/(2√(1 − 𝑥)) = (2 − 3𝑥)/(2√(1 − 𝑥)) Putting f’(𝒙)=𝟎 (2 − 3𝑥)/(2√(1 − 𝑥))=0 2 – 3𝑥 = 0 × 2√(1−𝑥) 2 – 3𝑥=0 – 3𝑥=−2 𝑥 =2/3 Finding f’’(𝒙) f’(𝑥)=(2 − 3𝑥)/(2√(1 − 𝑥)) f’’(𝑥)=𝑑/𝑑𝑥 ((2 − 3𝑥)/(2√(1 − 𝑥))) = 1/2 [(𝑑(2 − 3𝑥)/𝑑𝑥 . √(1 − 𝑥) − 𝑑(√(1 − 𝑥))/𝑑𝑥 . (2 − 3𝑥))/(√(1 − 𝑥))^2 ] = 1/2 [((0 − 3) √(1 − 𝑥) − 1/(2√(1 − 𝑥)) . 𝑑(1 − 𝑥)/𝑑𝑥 . (2 − 3𝑥))/((1 − 𝑥) )] = 1/2 [(−3√(1 − 𝑥) − 1/(2√(1 − 𝑥)) (0 − 1) . (2 − 3𝑥))/((1 − 𝑥) )] = 1/2 [(−3√(1 − 𝑥) + (2 − 3𝑥)/(2√(1 − 𝑥)) )/(1 − 𝑥)] = 1/2 [((−3√(1 − 𝑥)) (2√(1 − 𝑥)) + 2 − 3𝑥 )/(2(1 − 𝑥) √(1 − 𝑥))] = 1/2 [(−6(1 − 𝑥) + 2 − 3𝑥 )/(2(1 − 𝑥) √(1 − 𝑥))] = 1/2 [(−6 + 6𝑥 + 2 − 3𝑥 )/(2(1 − 𝑥) √(1 − 𝑥))] = 1/4 [(−4 + 3𝑥 )/(1 + 𝑥)^(3/2) ] Hence, f’’(𝑥)=1/4 [(−4 + 3𝑥 )/(1 + 𝑥)^(3/2) ] Putting 𝑥=2/3 f’’(2/3)=1/4 [(−4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(−4 + 2)/(5/3)^(3/2) ] =1/4 [(−2)/(5/3)^(3/2) ] = (− 1)/2 (3/5)^(3/2) < 0 Since f’’(𝑥)<0 when 𝑥 = 2/3 Hence, 𝑥=2/3 is the maxima Finding Maximum value of f(𝒙)=𝒙√(𝟏−𝒙) Putting 𝑥=2/3 f(2/3)=2/3 √(1−2/3) =2/3 √((3 − 2)/3) Finding Maximum value of f(𝒙)=𝒙√(𝟏−𝒙) Putting 𝑥=2/3 f(𝑥) = 𝑥√(1−𝑥) f(2/3)=2/3 √(1−2/3) =2/3 √((3 − 2)/3) =2/3 √(1/3) =2/(3√3) =2/(3√3) × √3/√3 =(2√3)/9 Maximum value of f(𝒙) is (𝟐√𝟑)/𝟗 at x = 𝟐/𝟑