Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
























Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐ฅ) = (2๐ฅ โ 1)^2 + 3 f(๐ฅ)=(2๐ฅโ1)^2+3 Hence, Minimum value of (2๐ฅโ1)^2 = 0 Minimum value of (2๐ฅโ1^2 )+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 ๐ฅ โด 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 (๐ฅ) = (2๐ฅ โ 1)^2+3 Finding fโ(x) f(๐ฅ)=(2๐ฅโ1)^2+3 fโ(๐ฅ)= 2(2๐ฅโ1) Putting fโ(๐)=๐ 2(2๐ฅโ1)=0 2๐ฅ โ 1 = 0 2๐ฅ = 1 ๐ฅ = 1/2 Thus, x = 1/2 is the minima Finding minimum value f(๐ฅ)=(2๐ฅโ1)^2+3 Putting ๐ฅ = 1/2 f(1/2)=(2 ร 1/2โ1)^2+3= (1โ1)^2+3= 3 โด 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 (๐ฅ)= (2๐ฅ โ 1)^2 + 3 Double Derivative Test f(๐ฅ)=(2๐ฅโ1)^2+3 Finding fโ(๐) fโ(๐ฅ)=2(2๐ฅโ1)^(2โ1) = 2(2๐ฅโ1) Putting fโ(๐)=๐ 2(2๐ฅโ1)=0 (2๐ฅโ1)=0 2๐ฅ = 0 + 1 ๐ฅ = 1/2 Finding fโโ(๐) fโ(๐ฅ)=2(2๐ฅโ1) fโ(๐ฅ) = 4๐ฅ โ 2 fโโ(๐ฅ)= 4 So, fโโ (1/2) = 4 Since fโโ (1/2) > 0 , ๐ฅ = 1/2 is point of local minima Putting ๐ฅ = 1/2 , we can calculate minimum value f(๐ฅ) = (2๐ฅโ1)^2+3 f(1/2)= (2(1/2)โ1)^2+3= (1โ1)^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 (๐ฅ) = 9๐ฅ2+12๐ฅ+2 Finding fโ(x) f (๐ฅ)=9๐ฅ2+12๐ฅ+2 Diff. w.r.t ๐ฅ fโ(๐ฅ)=18๐ฅ+12 fโ(๐ฅ)=6(3๐ฅ+2) Putting fโ(๐)=๐ 6(3๐ฅ+2)=0 3๐ฅ+2=0 3๐ฅ=โ2 ๐ฅ=(โ2)/( 3) Hence ๐ฅ=(โ2)/3 is point of minima of f(๐ฅ) Finding minimum value of f(๐ฅ) at ๐ฅ=(โ2)/3 f(๐ฅ)=9๐ฅ^2+12๐ฅ+2 Putting ๐ฅ=(โ2)/3 f(๐ฅ)=9((โ2)/3)^2+12((โ2)/3)+2=9(4/3)โ12(2/3)+2=โ2 Thus, Minimum value of f(๐)=โ๐ 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 (๐ฅ) = 9๐ฅ2+12๐ฅ+2 Finding fโ(๐) f (๐ฅ)=9๐ฅ2+12๐ฅ+2 Diff. w.r.t ๐ฅ fโ(๐ฅ)=๐(9๐ฅ^2 + 12๐ฅ + 2)/๐๐ฅ fโ(๐ฅ)=18๐ฅ+12 fโ(๐ฅ)=6(3๐ฅ+2) Putting fโ(๐)=๐ 6(3๐ฅ+2)=0 3๐ฅ+2=0 3๐ฅ=โ2 ๐ฅ=(โ2)/3 Finding fโโ(๐) fโ(๐ฅ)= 6(3๐ฅ+2) Again diff w.r.t ๐ฅ fโโ(๐ฅ)=๐(6(3๐ฅ+2))/๐๐ฅ fโโ(๐ฅ)=6 ๐(3๐ฅ + 2)/๐๐ฅ fโโ(๐ฅ)=6(3+0) fโโ(๐ฅ)=6(3) fโโ(๐ฅ)=18 So, fโโ((โ2)/3)=18 Since fโโ(๐ฅ)>0 is for ๐ฅ=(โ2)/3 ๐ฅ=(โ2)/3 is point of local minima Finding minimum value Putting ๐ฅ=(โ2)/3 in f(๐ฅ) f (๐ฅ)=9๐ฅ2+12๐ฅ+2 f ((โ2)/3)=9((โ2)/3)^2+12((โ2)/3)+2 =9(4/9)+12((โ2)/3)+2 =4โ8+2 =โ2 Hence, minimum value = โ2 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 (๐ฅ) = โ(๐ฅ โ 1)^2+10 f(๐ฅ)=โ(๐ฅโ1)^2+10 Finding fโ (x) Diff w.r.t ๐ฅ fโ(๐ฅ)=๐(โ(๐ฅโ1)^2+10)/๐๐ฅ fโ(๐ฅ) = โ2(๐ฅโ1)(๐(๐ฅโ1)/๐๐ฅ)+0 fโ(๐ฅ) = โ2(๐ฅโ1)(1โ0) + 0 fโ(๐ฅ)=โ2(๐ฅโ1) Putting fโ(๐)=๐ โ2(๐ฅโ1)=0 (๐ฅโ1)=0 ๐ฅ=1 Hence, ๐ฅ=1 is point of Maxima & No point of Minima Thus, f(๐ฅ) has maximum value at ๐ฅ=1 Putting x=1 in f(x) f(๐ฅ)=โ(๐ฅโ1)^2+10 f(1)=โ(1โ1)^2+10 = 0 + 10 = 10 Maximum value of f(๐ฅ) 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 (๐ฅ) = โ (๐ฅ โ 1)^2+10 f(๐ฅ)=โ(๐ฅโ1)^2+10 Finding fโ(x) Diff w.r.t ๐ฅ fโ(๐ฅ)=๐(โ(๐ฅโ1)^2+10)/๐๐ฅ fโ(๐ฅ) = โ2(๐ฅโ1)(๐(๐ฅโ1)/๐๐ฅ)+0 fโ(๐ฅ) = โ2(๐ฅโ1)(1โ0) + 0 fโ(๐ฅ)=โ2(๐ฅโ1) Putting fโ(๐)=๐ โ2(๐ฅโ1)=0 (๐ฅโ1)=0 ๐ฅ=1 Finding fโโ(๐) fโ(๐ฅ)=โ2(๐ฅโ1) Again diff w.r.t ๐ฅ fโโ(๐ฅ)=๐(โ2(๐ฅ โ 1))/๐๐ฅ =โ2 ๐(๐ฅ โ 1)/๐๐ฅ =โ2(1โ0) =โ2 Since fโโ(๐ฅ) < 0 for ๐ฅ=1 Hence, f(๐ฅ) has Maximum value at ๐ฅ=1 Finding maximum value of f(๐) f(๐ฅ)=โ(๐ฅโ1)^2+10 Putting ๐ฅ=1 f(๐ฅ) =โ(1โ1)^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(๐ฅ) = ๐ฅ3 + 1 f(๐ฅ)=๐ฅ^3+1 Finding fโ(x) fโ(๐ฅ)=๐(๐ฅ^3 + 1)/๐๐ฅ =3๐ฅ^2 Putting fโ(๐)=๐ 3๐ฅ^2=0 ๐ฅ^2=0 ๐ฅ=0 Therefore by first derivate test, the point ๐ฅ=0 is neither a point of local maxima nor a point of local Minima Hence ๐=๐ 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(๐ฅ) = ๐ฅ3 + 1 f(๐ฅ)=๐ฅ^3+1 Finding fโ(x) fโ(๐ฅ)=๐(๐ฅ^3+1)/๐๐ฅ =3๐ฅ^2 Putting fโ(๐)=๐ 3๐ฅ^2=0 ๐ฅ^2=0 f(๐ฅ)=๐ฅ^3+1 Finding fโ(x) fโ(๐ฅ)=๐(๐ฅ^3+1)/๐๐ฅ =3๐ฅ^2 Putting fโ(๐)=๐ 3๐ฅ^2=0 ๐ฅ^2=0
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