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

Transcript

Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐‘ฅ) = (2๐‘ฅ โ€“ 1)^2 + 3 f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Hence, Minimum value of (2๐‘ฅโˆ’1)^2 = 0 Minimum value of (2๐‘ฅโˆ’1^2 )+3 = 0 + 3 = 3 Square of number cant be negative It can be 0 or greater than 0 Also, there is no maximum value of ๐‘ฅ โˆด There is no maximum value of f(x) Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐‘ฅ) = (2๐‘ฅ โ€“ 1)^2+3 Finding fโ€™(x) f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 fโ€™(๐‘ฅ)= 2(2๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ 2(2๐‘ฅโˆ’1)=0 2๐‘ฅ โ€“ 1 = 0 2๐‘ฅ = 1 ๐‘ฅ = 1/2 Thus, x = 1/2 is the minima Finding minimum value f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Putting ๐‘ฅ = 1/2 f(1/2)=(2 ร— 1/2โˆ’1)^2+3= (1โˆ’1)^2+3= 3 โˆด Minimum value = 3 There is no maximum value Ex 6.5, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) ๐‘“ (๐‘ฅ)= (2๐‘ฅ โ€“ 1)^2 + 3 Double Derivative Test f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=2(2๐‘ฅโˆ’1)^(2โˆ’1) = 2(2๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ 2(2๐‘ฅโˆ’1)=0 (2๐‘ฅโˆ’1)=0 2๐‘ฅ = 0 + 1 ๐‘ฅ = 1/2 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=2(2๐‘ฅโˆ’1) fโ€™(๐‘ฅ) = 4๐‘ฅ โ€“ 2 fโ€™โ€™(๐‘ฅ)= 4 So, fโ€™โ€™ (1/2) = 4 Since fโ€™โ€™ (1/2) > 0 , ๐‘ฅ = 1/2 is point of local minima Putting ๐‘ฅ = 1/2 , we can calculate minimum value f(๐‘ฅ) = (2๐‘ฅโˆ’1)^2+3 f(1/2)= (2(1/2)โˆ’1)^2+3= (1โˆ’1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (๐‘ฅ) = 9๐‘ฅ2+12๐‘ฅ+2 Finding fโ€™(x) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 Diff. w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=18๐‘ฅ+12 fโ€™(๐‘ฅ)=6(3๐‘ฅ+2) Putting fโ€™(๐’™)=๐ŸŽ 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/( 3) Hence ๐‘ฅ=(โˆ’2)/3 is point of minima of f(๐‘ฅ) Finding minimum value of f(๐‘ฅ) at ๐‘ฅ=(โˆ’2)/3 f(๐‘ฅ)=9๐‘ฅ^2+12๐‘ฅ+2 Putting ๐‘ฅ=(โˆ’2)/3 f(๐‘ฅ)=9((โˆ’2)/3)^2+12((โˆ’2)/3)+2=9(4/3)โˆ’12(2/3)+2=โˆ’2 Thus, Minimum value of f(๐’™)=โˆ’๐Ÿ There is no maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (๐‘ฅ) = 9๐‘ฅ2+12๐‘ฅ+2 Finding fโ€™(๐’™) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 Diff. w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(9๐‘ฅ^2 + 12๐‘ฅ + 2)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=18๐‘ฅ+12 fโ€™(๐‘ฅ)=6(3๐‘ฅ+2) Putting fโ€™(๐’™)=๐ŸŽ 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/3 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)= 6(3๐‘ฅ+2) Again diff w.r.t ๐‘ฅ fโ€™โ€™(๐‘ฅ)=๐‘‘(6(3๐‘ฅ+2))/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6 ๐‘‘(3๐‘ฅ + 2)/๐‘‘๐‘ฅ fโ€™โ€™(๐‘ฅ)=6(3+0) fโ€™โ€™(๐‘ฅ)=6(3) fโ€™โ€™(๐‘ฅ)=18 So, fโ€™โ€™((โˆ’2)/3)=18 Since fโ€™โ€™(๐‘ฅ)>0 is for ๐‘ฅ=(โˆ’2)/3 ๐‘ฅ=(โˆ’2)/3 is point of local minima Finding minimum value Putting ๐‘ฅ=(โˆ’2)/3 in f(๐‘ฅ) f (๐‘ฅ)=9๐‘ฅ2+12๐‘ฅ+2 f ((โˆ’2)/3)=9((โˆ’2)/3)^2+12((โˆ’2)/3)+2 =9(4/9)+12((โˆ’2)/3)+2 =4โˆ’8+2 =โˆ’2 Hence, minimum value = โ€“2 There is no maximum value Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐‘“ (๐‘ฅ) = โ€“(๐‘ฅ โ€“ 1)^2+10 f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Finding fโ€™ (x) Diff w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(โˆ’(๐‘ฅโˆ’1)^2+10)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(๐‘‘(๐‘ฅโˆ’1)/๐‘‘๐‘ฅ)+0 fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(1โˆ’0) + 0 fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Hence, ๐‘ฅ=1 is point of Maxima & No point of Minima Thus, f(๐‘ฅ) has maximum value at ๐‘ฅ=1 Putting x=1 in f(x) f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 f(1)=โˆ’(1โˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(๐‘ฅ) is 10 There is no minimum value of f(๐’™) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐‘“ (๐‘ฅ) = โ€“ (๐‘ฅ โ€“ 1)^2+10 f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Finding fโ€™(x) Diff w.r.t ๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(โˆ’(๐‘ฅโˆ’1)^2+10)/๐‘‘๐‘ฅ fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(๐‘‘(๐‘ฅโˆ’1)/๐‘‘๐‘ฅ)+0 fโ€™(๐‘ฅ) = โ€“2(๐‘ฅโˆ’1)(1โˆ’0) + 0 fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=โˆ’2(๐‘ฅโˆ’1) Again diff w.r.t ๐‘ฅ fโ€™โ€™(๐‘ฅ)=๐‘‘(โˆ’2(๐‘ฅ โˆ’ 1))/๐‘‘๐‘ฅ =โˆ’2 ๐‘‘(๐‘ฅ โˆ’ 1)/๐‘‘๐‘ฅ =โˆ’2(1โˆ’0) =โˆ’2 Since fโ€™โ€™(๐‘ฅ) < 0 for ๐‘ฅ=1 Hence, f(๐‘ฅ) has Maximum value at ๐‘ฅ=1 Finding maximum value of f(๐’™) f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Putting ๐‘ฅ=1 f(๐‘ฅ) =โˆ’(1โˆ’1)^2+10 = 0 + 10 = 10 Maximum value of f(๐’™) is 10 There is no minimum value of f(๐’™) Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iv) f(๐‘ฅ) = ๐‘ฅ3 + 1 f(๐‘ฅ)=๐‘ฅ^3+1 Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3 + 1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Putting fโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2=0 ๐‘ฅ^2=0 ๐‘ฅ=0 Therefore by first derivate test, the point ๐‘ฅ=0 is neither a point of local maxima nor a point of local Minima Hence ๐’™=๐ŸŽ is point of inflexion Hence, there is no minimum or maximum value Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iv) f(๐‘ฅ) = ๐‘ฅ3 + 1 f(๐‘ฅ)=๐‘ฅ^3+1 Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3+1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Putting fโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2=0 ๐‘ฅ^2=0 f(๐‘ฅ)=๐‘ฅ^3+1 Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3+1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Putting fโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2=0 ๐‘ฅ^2=0

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.