Ex 6.5, 1 - Find maximum and minimum values of (i) f(x) = - Ex 6.5

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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 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 Step 1: Finding fโ€™(x) f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 fโ€™(๐‘ฅ)= 2(2๐‘ฅโˆ’1) Step 2: Putting fโ€™(๐‘ฅ)=0 2(2๐‘ฅโˆ’1)=0 2๐‘ฅ โ€“ 1 = 0 2๐‘ฅ = 1 ๐‘ฅ = 1/2 Step 3: 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 fโ€™(๐‘ฅ)=2(2๐‘ฅโˆ’1)^(2โˆ’1) = 2(2๐‘ฅโˆ’1) Putting fโ€™(๐‘ฅ)=0 2(2๐‘ฅโˆ’1)=0 (2๐‘ฅโˆ’1)=0 2๐‘ฅ = 0 + 1 ๐‘ฅ = 1/2 Now, finding fโ€™โ€™(๐‘ฅ) 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 Step 1: 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) Step 2: Putting fโ€™(๐‘ฅ)=0 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/( 3) Step 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 Step 1: 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) Step 2: Putting fโ€™(๐‘ฅ)=0 6(3๐‘ฅ+2)=0 3๐‘ฅ+2=0 3๐‘ฅ=โˆ’2 ๐‘ฅ=(โˆ’2)/3 Step 3: Finding fโ€™โ€™(๐‘ฅ) 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 Step 4: Putting ๐‘ฅ=(โˆ’2)/3 we can calculate minimum value of 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 Step 1: 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) Step 2: Putting fโ€™(๐‘ฅ)=0 โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Step 3: Hence, ๐‘ฅ=1 is point of Maxima & No point of Minima Thus, f(๐‘ฅ) has maximum value at ๐‘ฅ=1 f(๐‘ฅ)=โˆ’(๐‘ฅโˆ’1)^2+10 Putting ๐‘ฅ=1 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 Step 1: 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) Step 2: Putting fโ€™(๐‘ฅ)=0 โ€“2(๐‘ฅโˆ’1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Step 3: Finding fโ€™โ€™(๐‘ฅ) 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‘ฅ) 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 Step 1: Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3+1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Step 2: Putting fโ€™(๐‘ฅ)=0 3๐‘ฅ^2=0 ๐‘ฅ^2=0 ๐‘ฅ=0 Step 3: โ‡’ 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 Step 1: Finding fโ€™(x) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3+1)/๐‘‘๐‘ฅ =3๐‘ฅ^2 Step 2: Putting fโ€™(๐‘ฅ)=0 3๐‘ฅ^2=0 ๐‘ฅ^2=0 ๐‘ฅ=0 Step 3: Finding fโ€™โ€™(x) fโ€™(x) = 3x2 fโ€™โ€™(x) = 6x Finding fโ€™โ€™(x) at x = 0 fโ€™โ€™(0) = 6 ร— 0 = 0 Since fโ€™โ€™(x) = 0 at x = 0 โ‡’ 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

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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.