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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 28

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 29
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 30 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 31 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 32 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 33 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 34 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 35

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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‘₯√(1βˆ’π‘₯), π‘₯ > 0Finding f’(𝒙) f’(π‘₯)=𝑑(π‘₯√(1 βˆ’ π‘₯))/𝑑π‘₯ f’(π‘₯)=𝑑(π‘₯)/𝑑π‘₯ . √(1βˆ’π‘₯) + 𝑑(√(1 βˆ’ π‘₯))/𝑑π‘₯ . π‘₯ = 1 . √(1βˆ’π‘₯) + 1/(2√(1 βˆ’ π‘₯)) . 𝑑(1 βˆ’ π‘₯)/𝑑π‘₯ . π‘₯ = √(1βˆ’π‘₯) + 1/(2√(1 βˆ’ π‘₯)) (0 βˆ’1) . π‘₯ = √(1βˆ’π‘₯) – π‘₯/(2√(1 βˆ’ π‘₯)) = (2(√(1 βˆ’ π‘₯) )^2βˆ’ π‘₯)/(2√(1 βˆ’ π‘₯)) = (2(1 βˆ’ π‘₯) βˆ’ π‘₯)/(2√(1 βˆ’ π‘₯)) = (2 βˆ’ 2π‘₯ βˆ’ π‘₯)/(2√(1 βˆ’ π‘₯)) = (2 βˆ’ 3π‘₯)/(2√(1 βˆ’ π‘₯)) Putting f’(𝒙)=𝟎 (2 βˆ’ 3π‘₯)/(2√(1 βˆ’ π‘₯))=0 2 – 3π‘₯ = 0 Γ— 2√(1βˆ’π‘₯) 2 – 3π‘₯=0 – 3π‘₯=βˆ’2 π‘₯ =2/3 Finding f’’(𝒙) f’(π‘₯)=(2 βˆ’ 3π‘₯)/(2√(1 βˆ’ π‘₯)) f’’(π‘₯)=𝑑/𝑑π‘₯ ((2 βˆ’ 3π‘₯)/(2√(1 βˆ’ π‘₯))) = 1/2 [(𝑑(2 βˆ’ 3π‘₯)/𝑑π‘₯ . √(1 βˆ’ π‘₯) βˆ’ 𝑑(√(1 βˆ’ π‘₯))/𝑑π‘₯ . (2 βˆ’ 3π‘₯))/(√(1 βˆ’ π‘₯))^2 ] = 1/2 [((0 βˆ’ 3) √(1 βˆ’ π‘₯) βˆ’ 1/(2√(1 βˆ’ π‘₯)) . 𝑑(1 βˆ’ π‘₯)/𝑑π‘₯ . (2 βˆ’ 3π‘₯))/((1 βˆ’ π‘₯) )] Using quotient rule as (𝑒/𝑣)^β€²=(𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 = 1/2 [(βˆ’3√(1 βˆ’ π‘₯) βˆ’ 1/(2√(1 βˆ’ π‘₯)) (0 βˆ’ 1) . (2 βˆ’ 3π‘₯))/((1 βˆ’ π‘₯) )] = 1/2 [(βˆ’3√(1 βˆ’ π‘₯) + (2 βˆ’ 3π‘₯)/(2√(1 βˆ’ π‘₯)) )/(1 βˆ’ π‘₯)] = 1/2 [((βˆ’3√(1 βˆ’ π‘₯)) (2√(1 βˆ’ π‘₯)) + 2 βˆ’ 3π‘₯ )/(2(1 βˆ’ π‘₯) √(1 βˆ’ π‘₯))] = 1/2 [(βˆ’6(1 βˆ’ π‘₯) + 2 βˆ’ 3π‘₯ )/(2(1 βˆ’ π‘₯) √(1 βˆ’ π‘₯))] = 1/2 [(βˆ’6 + 6π‘₯ + 2 βˆ’ 3π‘₯ )/(2(1 βˆ’ π‘₯) √(1 βˆ’ π‘₯))] = 1/4 [(βˆ’4 + 3π‘₯ )/(1 + π‘₯)^(3/2) ] Hence, f’’(π‘₯)=1/4 [(βˆ’4 + 3π‘₯ )/(1 + π‘₯)^(3/2) ] Putting π‘₯=2/3 f’’(2/3)=1/4 [(βˆ’4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(βˆ’4 + 2)/(5/3)^(3/2) ] =1/4 [(βˆ’2)/(5/3)^(3/2) ] = (βˆ’ 1)/2 (3/5)^(3/2) < 0 Since f’’(π‘₯)<0 when π‘₯ = 2/3 Hence, π‘₯=2/3 is the maxima Finding Maximum value of f(𝒙)=π’™βˆš(πŸβˆ’π’™) Putting π‘₯=2/3 f(2/3)=2/3 √(1βˆ’2/3) =2/3 √((3 βˆ’ 2)/3) Finding Maximum value of f(𝒙)=π’™βˆš(πŸβˆ’π’™) Putting π‘₯=2/3 f(π‘₯) = π‘₯√(1βˆ’π‘₯) f(2/3)=2/3 √(1βˆ’2/3) =2/3 √((3 βˆ’ 2)/3) =2/3 √(1/3) =2/(3√3) =2/(3√3) Γ— √3/√3 =(2√3)/9 Maximum value of f(𝒙) is (πŸβˆšπŸ‘)/πŸ— at x = 𝟐/πŸ‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.