Ex 6.5, 3 - Find local maxima, local minima of (i) f(x) = x2

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 5

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 8 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 9 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 10 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 11 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 12 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 13 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 14 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 15 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 16 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 17 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 18 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 19 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 20 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 21 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 22 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 23 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 24 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 25 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 26 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 27 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 & the local minimum values, as the case may be: (i) f (π‘₯)=π‘₯2 f(π‘₯)=π‘₯^2 Finding f’(x) f’(x) = 2x Putting f’(x) = 0 2x = 0 x = 0 Finding f’’(x) f’(x) = 2x Differentiating again f’’(x) = 2 Since f’’(x) > 0 for x = 0 So, f(x) is minimum at x = 0 Minimum value of f(x) at x = 0 f(x) = x2 f(0) = 02 = 0 So, Minimum value of f(x) = 0 There is no maximum value 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: (ii) 𝑔(π‘₯)=π‘₯3 –3π‘₯𝑔(π‘₯)=π‘₯3 –3π‘₯ Finding g’(𝒙) g’(π‘₯)=𝑑(π‘₯^3 βˆ’ 3π‘₯)/𝑑π‘₯ g’(π‘₯)=3π‘₯^2βˆ’3 Putting g’(𝒙)=𝟎 3π‘₯^2βˆ’3=0 3π‘₯^2=3 π‘₯^2=3/3 π‘₯^2=1 π‘₯=Β±1 So, x = 1 & x = –1 Finding g’’(𝒙) g’(π‘₯)=3π‘₯^2βˆ’3 g’’(π‘₯)=𝑑(3π‘₯^2βˆ’3)/𝑑π‘₯ = 6π‘₯βˆ’0 = 6π‘₯ Putting 𝒙=𝟏 in g’’(x) g’’(1)=6(1)= 6 > 0 Thus, g’’(π‘₯)>0 when π‘₯=1 β‡’ π‘₯=1 is point of local minima & g(π‘₯) is minimum at π‘₯=1 Local minimum value g(π‘₯)=π‘₯^3βˆ’3π‘₯ g(1)=(1)^3βˆ’3(1) =1βˆ’3 =βˆ’πŸ Putting 𝒙=βˆ’πŸ in g’’(x) g’’(βˆ’1)=6(βˆ’1)= –6 < 0 Thus, g’’(π‘₯)<0 when π‘₯=βˆ’1 β‡’ π‘₯=βˆ’1 is point of local maxima & g(π‘₯) is maximum at π‘₯=βˆ’1 Local minimum value g(π‘₯)=π‘₯^3βˆ’3π‘₯ g(βˆ’1)=(βˆ’1)^3βˆ’3(βˆ’1) =βˆ’1+3 =𝟐 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: (iii) β„Ž(π‘₯)=sin⁑π‘₯+cos⁑π‘₯, 0<π‘₯<πœ‹/2 β„Ž(π‘₯)=sin⁑π‘₯+cos⁑π‘₯, 0<π‘₯<πœ‹/2 Finding 𝒉′(𝒙) β„Žβ€²(π‘₯)=𝑑(sin⁑π‘₯ + cos⁑π‘₯" " )/𝑑π‘₯ β„Ž^β€² (π‘₯)=cos⁑π‘₯βˆ’sin⁑π‘₯ Putting 𝒉′(𝒙)=𝟎 cos⁑π‘₯βˆ’π‘ π‘–π‘›π‘₯=0 cos⁑〖π‘₯=𝑠𝑖𝑛 π‘₯γ€— 1 = sin⁑π‘₯/cos⁑π‘₯ 1 = tan π‘₯ tan π‘₯=1 ∴ π‘₯=45Β°= πœ‹/4 Finding h’’(𝒙) h’(π‘₯)=cos⁑π‘₯βˆ’sin⁑π‘₯ h’’(π‘₯)=βˆ’sin⁑π‘₯βˆ’cos⁑π‘₯ Putting 𝒙=𝝅/πŸ’ h’’(Ο€/4)=βˆ’sin(Ο€/4)βˆ’π‘π‘œπ‘ (Ο€/4) = – 1/√2βˆ’1/√2 = (βˆ’2)/√2 = – √2 Since h’’(π‘₯)<0 when π‘₯=Ο€/4 ∴ π‘₯=Ο€/4 is point of Local Maxima f has Maximum value at 𝒙=𝝅/πŸ’ f(π‘₯)=𝑠𝑖𝑛π‘₯+π‘π‘œπ‘ π‘₯ f(Ο€/4)=𝑠𝑖𝑛(Ο€/4)+π‘π‘œπ‘ (Ο€/4) = 1/√2+1/√2 = 2/√2 = √𝟐 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: (iv) f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹f (π‘₯)=sin⁑π‘₯ –cos⁑π‘₯, 0<π‘₯<2 πœ‹ Finding f’(𝒙) f’(π‘₯)=cos⁑π‘₯βˆ’(βˆ’sin⁑π‘₯ ) f’(π‘₯)=cos⁑π‘₯+sin⁑π‘₯ Putting f’(𝒙)=𝟎 cos⁑π‘₯+sin⁑π‘₯ = 0 cos⁑π‘₯=βˆ’sin⁑π‘₯ 1=(βˆ’sin⁑π‘₯)/cos⁑π‘₯ (βˆ’sin⁑π‘₯)/cos⁑π‘₯ =1 – tan π‘₯=1 tan π‘₯=βˆ’1 Since 0 < π‘₯ < 2Ο€ & tan π‘₯ is negative tan ΞΈ lies in either 2nd or 4th quadrant So, value of π‘₯ is π‘₯=3πœ‹/4 π‘œπ‘Ÿ 7πœ‹/4 Now finding f’’(π‘₯) f’’(π‘₯)=𝑑(cos⁑π‘₯ + sin⁑π‘₯ )/𝑑π‘₯ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting 𝒙 = πŸ‘π…/πŸ’ f’’(3πœ‹/4)=βˆ’π‘ π‘–π‘›(3πœ‹/4)+π‘π‘œπ‘ (3πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (πœ‹βˆ’ πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)+(βˆ’π‘π‘œπ‘  πœ‹/4) =(βˆ’1)/√2βˆ’1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 < 0 Hence f’’(π‘₯)<0 when π‘₯ = 3πœ‹/4 Thus π‘₯ = 3πœ‹/4 is point of local maxima ∴ f(π‘₯) is maximum value at π‘₯ = 3πœ‹/4 The local maximum value is f(π‘₯)=sin⁑π‘₯βˆ’cos⁑π‘₯ f(3πœ‹/4)=𝑠𝑖𝑛(3πœ‹/4)βˆ’π‘π‘œπ‘ (3πœ‹/4) =𝑠𝑖𝑛(πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹βˆ’πœ‹/4) =𝑠𝑖𝑛(πœ‹/4)βˆ’(βˆ’π‘π‘œπ‘  πœ‹/4) As sin (180 βˆ’ πœƒ) = sin πœƒ & cos (180 βˆ’ πœƒ) = βˆ’cos πœƒ =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2+1/√2 =2/√2 =√2 Now, for 𝒙 = πŸ•π…/πŸ’ f’’(π‘₯)=βˆ’sin⁑π‘₯+cos⁑π‘₯ Putting π‘₯ = 7πœ‹/4 f’’(7πœ‹/4)=βˆ’sin⁑(7πœ‹/4)+cos⁑(7πœ‹/4) f’’(7πœ‹/4)=βˆ’π‘ π‘–π‘›(2πœ‹βˆ’πœ‹/4)+π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’(βˆ’π‘ π‘–π‘›(πœ‹/4))+π‘π‘œπ‘ (πœ‹/4) =𝑠𝑖𝑛 πœ‹/4+π‘π‘œπ‘  πœ‹/4 =1/√2 + 1/√2 =2/√2 =√2 > 0 f’’(π‘₯)>0 when π‘₯ = 7πœ‹/4 Thus π‘₯ = 7πœ‹/4 is point of local minima f(π‘₯) has minimum value at π‘₯ = 7πœ‹/4 As sin (2Ο€ βˆ’ πœƒ) = –sin πœƒ & cos (2Ο€ βˆ’ πœƒ) = cos πœƒ Local minimum value is f(π‘₯)=𝑠𝑖𝑛π‘₯βˆ’π‘π‘œπ‘ π‘₯ f(7πœ‹/4)=𝑠𝑖𝑛(7πœ‹/4)βˆ’π‘π‘œπ‘ (7πœ‹/4) =𝑠𝑖𝑛(2πœ‹βˆ’πœ‹/4)βˆ’π‘π‘œπ‘ (2πœ‹βˆ’πœ‹/4) =βˆ’π‘ π‘–π‘›(πœ‹/4)βˆ’π‘π‘œπ‘ (πœ‹/4) =(βˆ’1)/√2 βˆ’ 1/√2 =(βˆ’2)/√2 =βˆ’βˆš2 Thus, f(π‘₯) is maximum at x = πŸ‘π…/πŸ’ and maximum value is √𝟐 & f(π‘₯) is minimum at x = πŸ•π…/πŸ’ and maximum value is β€“βˆšπŸ 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: (v) 𝑓 (π‘₯)=π‘₯3 –6π‘₯2+9π‘₯+15𝑓 (π‘₯)=π‘₯3 –6π‘₯2+9π‘₯+15 Finding f’(𝒙) f’(π‘₯)=𝑑(π‘₯3 – 6π‘₯2 + 9π‘₯ + 15" " )/𝑑π‘₯ f’(π‘₯)=3π‘₯^2βˆ’12π‘₯+9 f’(π‘₯)=3(π‘₯^2βˆ’4π‘₯+3) Putting f’(𝒙)=𝟎 3(π‘₯^2βˆ’4π‘₯+3)=0 π‘₯^2βˆ’4π‘₯+3=0 π‘₯^2βˆ’3π‘₯βˆ’π‘₯+3=0 π‘₯(π‘₯βˆ’3)βˆ’1(π‘₯βˆ’3)=0 (π‘₯βˆ’1)(π‘₯βˆ’3)=0 So, x = 1 & x = 3 Finding f’’(𝒙) f’(π‘₯)=3(π‘₯^2βˆ’4π‘₯+3) f’’(π‘₯)=𝑑(3(π‘₯^2 βˆ’ 4π‘₯+3))/𝑑π‘₯ = 3(2π‘₯βˆ’4+0) = 6π‘₯βˆ’12 Putting 𝒙=𝟏 in f’’(𝒙) f’’(1)=6(1)βˆ’12 = 6 – 12 = – 6 < 0 Since f’’(π‘₯)<0 when π‘₯=1 β‡’ π‘₯=1 is point of local maxima ∴ f(π‘₯) is maximum at 𝒙=𝟏 Maximum value of f(π‘₯) at π‘₯ = 1 f(π‘₯)=π‘₯^3βˆ’6π‘₯^2+9π‘₯+15 f(1)=(1)^3βˆ’6(1)^2+9(1)+15 = 1 – 6 + 9 + 15 = 19 Putting 𝒙=πŸ‘ in f’’(x) f’’(π‘₯)=6π‘₯βˆ’12 f’’(3)=6(3)βˆ’12 = 18 – 12 = 6 > 0 Since f’’(π‘₯)>0 when π‘₯=3 β‡’ π‘₯=3 is point of local minima ∴ f(π‘₯) is minimum at 𝒙=πŸ‘ Minimum value of f(π‘₯) at π‘₯ = 3 f(π‘₯)=π‘₯^3βˆ’6π‘₯^2+9π‘₯+15 f(3)=(3)^3βˆ’6(3)^2+9(3)+15 = 27 – 54 + 27 + 15 = 15 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: (vi) g (π‘₯) = π‘₯/2 + 2/π‘₯ , π‘₯ > 0g (π‘₯) = π‘₯/2 + 2/π‘₯, π‘₯ > 0 Finding g’(𝒙) g’(π‘₯)=𝑑/𝑑π‘₯ (π‘₯/2+2/π‘₯) =𝑑/𝑑π‘₯ (π‘₯/2)+𝑑/𝑑π‘₯ (2π‘₯^(βˆ’1) ) =1/2βˆ’2π‘₯^(βˆ’2) =1/2βˆ’2/π‘₯^2 Putting g’(𝒙)=𝟎 1/2βˆ’2/π‘₯^2 =0 (π‘₯^2βˆ’ 4)/(2π‘₯^2 )=0 π‘₯^2βˆ’4=0 Γ—2π‘₯^2 (π‘₯βˆ’2)(π‘₯+2)=0 So, π‘₯=2 & π‘₯=βˆ’2 Since π‘₯>0 is given, we consider only π‘₯=2 Finding g’’(𝒙) g’(π‘₯)=1/2βˆ’2/π‘₯^2 g’’(π‘₯)=𝑑/𝑑π‘₯ (1/2βˆ’2/π‘₯^2 ) = 0 – 2 . (βˆ’2) π‘₯^(βˆ’2βˆ’1) = 4π‘₯^(βˆ’3) = 4/π‘₯^3 Putting value π‘₯=2 in g’’(π‘₯) g’’(π‘₯)=4/(2)^3 = 4/8= 1/2 Since g’’ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(π‘₯) is g(π‘₯)=π‘₯/2+2/π‘₯ Putting π‘₯=2 g(2)= 2/2+2/2 = 1 + 1 = 2 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: (vii) g (π‘₯) = 1/(π‘₯^2 + 2)Finding g’(𝒙) g’(π‘₯)=𝑑/𝑑π‘₯ (1/(π‘₯^2 + 2)) g’(π‘₯)=(𝑑(π‘₯^2 + 2)^(βˆ’1))/𝑑π‘₯ g’(π‘₯)=βˆ’1(π‘₯^2+2)^(βˆ’1βˆ’1) Γ— (2π‘₯+0) g’(π‘₯)=βˆ’2π‘₯(π‘₯^2+2)^(βˆ’2) gβ€²(π‘₯)=( βˆ’2π‘₯ )/(π‘₯^2 + 2)^2 Putting g’(𝒙)=𝟎 ( βˆ’2π‘₯ )/(π‘₯^2+2)^2 =0 –2π‘₯=0 Γ—(π‘₯^2+2)^2 –2π‘₯=0 π‘₯=0 Finding g’’(𝒙) g’(π‘₯)=(βˆ’2π‘₯)/(π‘₯^2 + 2)^2 g’’(π‘₯)=(𝑑(βˆ’2π‘₯)/𝑑π‘₯ . γ€– (π‘₯^2 + 2)γ€—^2 βˆ’ (𝑑(π‘₯^2 + 2)^2)/𝑑π‘₯ . (βˆ’2π‘₯))/((π‘₯^2 + 2)^2 )^2 Using quotient rule as (𝑒/𝑣)^β€²=(𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)^(2βˆ’1).𝑑(π‘₯^2 + 2)/𝑑π‘₯ . (βˆ’2π‘₯))/((π‘₯^2 + 2)^2 )^2 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)(2π‘₯ + 0) (βˆ’2π‘₯))/(π‘₯^2 + 2)^4 =(βˆ’2 (π‘₯^2 + 2)^2βˆ’2 (π‘₯^2 + 2)(2π‘₯) (βˆ’2π‘₯))/(π‘₯^(2 )+ 2)^4 =(βˆ’2 (π‘₯^2 + 2)^2+ 8π‘₯^2 (π‘₯^2 + 2))/(π‘₯^(2 )+ 2)^4 =(βˆ’2 (π‘₯^2 + 2)[(π‘₯^(2 )+ 2) βˆ’ 4π‘₯^2 ])/(π‘₯^2 + 2)^4 =(βˆ’2 (π‘₯^2 + 2)(βˆ’3π‘₯^2 + 2))/(π‘₯^2 + 2)^4 =(βˆ’2(βˆ’3π‘₯^2 + 2))/(π‘₯^(2 )+ 2)^3 Putting x = 0 in g’’(x) g’’(0)=(βˆ’2(βˆ’3(0) + 2))/(0^2 + 2)^3 =(βˆ’2(0 + 2))/(2)^3 =(βˆ’4)/8=(βˆ’1)/2 Hence g’’(π‘₯)<0 when π‘₯ = 0 ∴ π‘₯ = 0 is point of local maxima Thus, g(π‘₯) is maximum at 𝒙 = 0 Maximum value of g(𝒙) at x = 0 g(π‘₯)=1/(π‘₯^(2 )+ 2) g(0)=1/(0^2 + 2) = 1/2 Maximum value is 𝟏/𝟐 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.