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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.