1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.5

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: (i) f (๐ฅ)=๐ฅ2 f(๐ฅ)=๐ฅ^2 First we plot the graph of ๐ฅ2 Note that :- At ๐ฅ = 0 , f(0)=0 Also, f(๐ฅ)>0 for all ๐ฅ , ๐ฅ โ R expect 0 Hence, x = 0 is point of minima of f (x) So, Minimum value of f(๐ฅ)=0 at ๐ฅ = 0 Since f(๐ฅ)>0 for ๐ฅ โ R expect 0 So, we can not find a maximum value e.g. f(100) =(100)^2= 10000 f(1000)= (1000)^2= 1000000 Hence , we cannot find a maximum value of f(๐ฅ) on , ๐ฅ โ R Hence no point of maximum value of f on R Ex 6.5,3 (Method 1) 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) ๐(๐ฅ)=๐ฅ3 โ3๐ฅ ๐(๐ฅ)=๐ฅ3 โ3๐ฅ Step 1: Finding gโ(๐ฅ) gโ(๐ฅ)=๐(๐ฅ^3โ3๐ฅ)/๐๐ฅ gโ(๐ฅ)=3๐ฅ^2โ3 Step 2: Putting gโ(๐ฅ)=0 3๐ฅ^2โ3=0 3๐ฅ^2=3 ๐ฅ^2=3/3 ๐ฅ^2=1 ๐ฅ=ยฑ1 So, x = 1 & x = โ1 Step 3: First derivative test Thus, โ ๐ฅ=1 is point of local minima โ g(๐ฅ) Is minimum at ๐ฅ=1 & gโ(๐ฅ) changes sign from (+)๐ฃ๐ to (โ)๐ฃ๐ when ๐ฅ=โ1 โ ๐ฅ=โ1 is point of local maxima โ g(๐ฅ) is maximum at ๐ฅ=โ1 Local minimum value of g(๐ฅ)=๐ฅ^3โ3๐ฅ at x = 1 g(1)=(1)^3โ3(1)=1โ3=โ๐ Local maximum value of g(๐ฅ)=๐ฅ^3โ3๐ฅ at x = โ1 g(โ1)= (โ1)^3โ3(โ1)=โ1+3 =๐ Ex 6.5,3 (Method 2) 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) ๐(๐ฅ)=๐ฅ3 โ3๐ฅ ๐(๐ฅ)=๐ฅ3 โ3๐ฅ Step 1: Finding gโ(๐ฅ) gโ(๐ฅ)=๐(๐ฅ^3โ3๐ฅ)/๐๐ฅ gโ(๐ฅ)=3๐ฅ^2โ3 Step 2: Putting gโ(๐ฅ)=0 3๐ฅ^2โ3=0 3๐ฅ^2=3 ๐ฅ^2=3/3 ๐ฅ^2=1 ๐ฅ=ยฑ1 So, x = 1 & x = โ1 Step 3: Finding gโโ(๐ฅ) gโ(๐ฅ)=3๐ฅ^2โ3 gโโ(๐ฅ)=๐(3๐ฅ^2โ3)/๐๐ฅ = 6๐ฅโ0 = 6๐ฅ Putting ๐=๐ in gโโ(x) gโโ(1)=6(1)= 6 > 0 Thus, gโโ(๐ฅ)>0 when ๐ฅ=1 โ ๐ฅ=1 is point of local minima & g(๐ฅ) is minimum at ๐ฅ=1 Local minimum value g(๐ฅ)=๐ฅ^3โ3๐ฅ g(1)=(1)^3โ3(1) =1โ3 =โ๐ Putting ๐=โ๐ in gโโ(x) gโโ(โ1)=6(โ1)= โ6 < 0 Thus, gโโ(๐ฅ)<0 when ๐ฅ=โ1 โ ๐ฅ=โ1 is point of local maxima & g(๐ฅ) is maximum at ๐ฅ=โ1 Local minimum value g(๐ฅ)=๐ฅ^3โ3๐ฅ g(โ1)=(โ1)^3โ3(โ1) =โ1+3 =๐ Ex 6.5,3 (Method 1) 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) โ(๐ฅ)=sinโก๐ฅ+cosโก๐ฅ, 0<๐ฅ<๐/2 โ(๐ฅ)=sinโก๐ฅ+cosโก๐ฅ, 0<๐ฅ<๐/2 Step 1: Finding โโฒ(๐ฅ) โโฒ(๐ฅ)=๐(sinโก๐ฅ+ cosโก๐ฅ" " )/๐๐ฅ โ^โฒ (๐ฅ)=cosโก๐ฅโsinโก๐ฅ Step 2: Putting โ(๐ฅ)=0 cosโก๐ฅโ๐ ๐๐๐ฅ=0 cosโกใ๐ฅ=๐ ๐๐ ๐ฅใ 1 = sinโก๐ฅ/cosโก๐ฅ 1 = tan ๐ฅ tan ๐ฅ=1 โด ๐ฅ=45ยฐ= ๐/4 Step 3: Since sign of fโ(x) changes from positive to negative, x = ๐/4 is Maxima Finding maximum value f has Maximum value at ๐ฅ=ฯ/4 f(๐ฅ)=๐ ๐๐๐ฅ+๐๐๐ ๐ฅ f(ฯ/4)=๐ ๐๐(ฯ/4)+๐๐๐ (ฯ/4) = 1/โ2+1/โ2 = 2/โ2 = โ๐ Ex 6.5,3 (Method 2) 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) โ(๐ฅ)=sinโก๐ฅ+cosโก๐ฅ, 0<๐ฅ<๐/2 โ(๐ฅ)=sinโก๐ฅ+cosโก๐ฅ, 0<๐ฅ<๐/2 Step 1: Finding โโฒ(๐ฅ) โโฒ(๐ฅ)=๐(sinโก๐ฅ+ cosโก๐ฅ" " )/๐๐ฅ โ^โฒ (๐ฅ)=cosโก๐ฅโsinโก๐ฅ Step 2: Putting โ(๐ฅ)=0 cosโก๐ฅโ๐ ๐๐๐ฅ=0 cosโกใ๐ฅ=๐ ๐๐ ๐ฅใ 1 = sinโก๐ฅ/cosโก๐ฅ 1 = tan ๐ฅ tan ๐ฅ=1 โด ๐ฅ=45ยฐ= ๐/4 Step 3: Finding hโโ(๐ฅ) hโ(๐ฅ)=๐๐๐ ๐ฅโ๐ ๐๐๐ฅ hโโ(๐ฅ)=โ๐ ๐๐๐ฅโ๐๐๐ ๐ฅ Putting ๐ฅ=ฯ/4 hโโ(ฯ/4)=โsin(ฯ/4)โ๐๐๐ (ฯ/4) = โ 1/โ2โ1/โ2 = (โ2)/โ2 = โ โ2 Since hโโ(๐ฅ)<0 when ๐ฅ=ฯ/4 โ๐ฅ=ฯ/4 is point of Local Maxima f has Maximum value at ๐=๐/๐ f(๐ฅ)=๐ ๐๐๐ฅ+๐๐๐ ๐ฅ f(ฯ/4)=๐ ๐๐(ฯ/4)+๐๐๐ (ฯ/4) = 1/โ2+1/โ2 = 2/โ2 = โ๐ 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 (๐ฅ)=sinโก๐ฅ โcosโก๐ฅ, 0<๐ฅ<2 ๐ f (๐ฅ)=sinโก๐ฅ โcosโก๐ฅ, 0<๐ฅ<2 ๐ Finding fโ(๐ฅ) fโ(๐ฅ)=cosโก๐ฅโ(โsinโก๐ฅ ) fโ(๐ฅ)=cosโก๐ฅ+sinโก๐ฅ Putting fโ(๐ฅ)=0 cosโก๐ฅ+sinโก๐ฅ = 0 cosโก๐ฅ=โsinโก๐ฅ 1=(โsinโก๐ฅ)/cosโก๐ฅ (โsinโก๐ฅ)/cosโก๐ฅ =1 โ tan ๐ฅ=1 tan ๐ฅ=โ1 Since 0 < ๐ฅ < 2ฯ & tan ๐ฅ is negative tan ฮธ lies in either (ii) or (iv) quadrant So, value of ๐ฅ is ๐ฅ=3๐/4 ๐๐ 7๐/4 Now finding fโโ(๐ฅ) fโโ(๐ฅ)=๐(cosโก๐ฅ +sinโก๐ฅ )/๐๐ฅ fโโ(๐ฅ)=โsinโก๐ฅ+cosโก๐ฅ Putting ๐ = ๐๐/๐ fโโ(3๐/4)=โ๐ ๐๐(3๐/4)+๐๐๐ (3๐/4) =โ๐ ๐๐(๐โ๐/4)+๐๐๐ (๐โ ๐/4) =โ๐ ๐๐(๐/4)+(โ๐๐๐  ๐/4) =(โ1)/โ2โ1/โ2 =(โ2)/โ2 =โโ2 < 0 Hence fโโ(๐ฅ)<0 when ๐ฅ = 3๐/4 Thus ๐ฅ = 3๐/4 is point of local maxima โด f(๐ฅ) is maximum value at ๐ฅ = 3๐/4 The local maximum value is f(๐ฅ)=sinโก๐ฅโcosโก๐ฅ f(3๐/4)=๐ ๐๐(3๐/4)โ๐๐๐ (3๐/4) =๐ ๐๐(๐โ๐/4)โ๐๐๐ (๐โ๐/4) =๐ ๐๐(๐/4)โ(โ๐๐๐  ๐/4) =๐ ๐๐ ๐/4+๐๐๐  ๐/4 =1/โ2+1/โ2 =2/โ2 =โ2 Now, for ๐ = ๐๐/๐ fโโ(๐ฅ)=โsinโก๐ฅ+cosโก๐ฅ Putting ๐ฅ = 7๐/4 fโโ(7๐/4)=โsinโก(7๐/4)+cosโก(7๐/4) fโโ(7๐/4)=โ๐ ๐๐(2๐โ๐/4)+๐๐๐ (2๐โ๐/4) =โ(โ๐ ๐๐(๐/4))+๐๐๐ (๐/4) =๐ ๐๐ ๐/4+๐๐๐  ๐/4 =1/โ2 + 1/โ2 =2/โ2 =โ2 > 0 fโโ(๐ฅ)>0 when ๐ฅ = 7๐/4 Thus ๐ฅ = 7๐/4 is point of local minima f(๐ฅ) has minimum value at ๐ฅ = 7๐/4 Local minimum value is f(๐ฅ)=๐ ๐๐๐ฅโ๐๐๐ ๐ฅ f(7๐/4)=๐ ๐๐(7๐/4)โ๐๐๐ (7๐/4) =๐ ๐๐(2๐โ๐/4)โ๐๐๐ (2๐โ๐/4) =โ๐ ๐๐(๐/4)โ๐๐๐ (๐/4) =(โ1)/โ2 โ 1/โ2 =(โ2)/โ2 =โโ2 Thus, f(๐ฅ) is maximum at x = ๐๐/๐ and maximum value is โ๐ & f(๐ฅ) is minimum at x = ๐๐/๐ and maximum value is โโ๐ Ex 6.5,3(Method 1) 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) ๐ (๐ฅ)=๐ฅ3 โ6๐ฅ2+9๐ฅ+15 ๐ (๐ฅ)=๐ฅ3 โ6๐ฅ2+9๐ฅ+15 Finding fโ(๐ฅ) fโ(๐ฅ)=๐(๐ฅ3 โ 6๐ฅ2+ 9๐ฅ + 15" " )/๐๐ฅ fโ(๐ฅ)=3๐ฅ^2โ12๐ฅ+9 fโ(๐ฅ)=3(๐ฅ^2โ4๐ฅ+3) Putting fโ(๐ฅ)=0 3(๐ฅ^2โ4๐ฅ+3)=0 ๐ฅ^2โ4๐ฅ+3=0 ๐ฅ^2โ3๐ฅโ๐ฅ+3=0 ๐ฅ(๐ฅโ3)โ1(๐ฅโ3)=0 (๐ฅโ1)(๐ฅโ3)=0 So, x = 1 & x = 3 Step 3: Using first derivative test at x = 1 & x = โ3 Thus, โ ๐ฅ = 1 is point of local maxima f(๐ฅ) has maximum value at ๐ฅ = 1 Thus, โ ๐ฅ = 1 is point of local maxima f(๐ฅ) has maximum value at ๐ฅ = 1 And, ๐ฅ = 3 is point of local minima โ f(๐ฅ) has minimum value at ๐ฅ = 3 Maximum value of f(๐ฅ) at ๐ = 1 f(๐ฅ)=๐ฅ^3โ6๐ฅ^2+9๐ฅ+15 f(๐ฅ)=(1)^3+6(1)^2+9(1)+15 = 1 โ 6 + 9 + 15 = 4 + 15 = 19 Minimum value of f(๐ฅ) at ๐ = 3 f(๐ฅ)=๐ฅ^3โ6๐ฅ^2+9๐ฅ+15 f(๐ฅ)=(3)^3โ6(3)^2+9(3)+15 = 27 โ 54 + 27 + 15 = 15 Ex 6.5,3(Method 2) 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) ๐ (๐ฅ)=๐ฅ3 โ6๐ฅ2+9๐ฅ+15 ๐ (๐ฅ)=๐ฅ3 โ6๐ฅ2+9๐ฅ+15 Finding fโ(๐ฅ) fโ(๐ฅ)=๐(๐ฅ3 โ 6๐ฅ2+ 9๐ฅ + 15" " )/๐๐ฅ fโ(๐ฅ)=3๐ฅ^2โ12๐ฅ+9 fโ(๐ฅ)=3(๐ฅ^2โ4๐ฅ+3) Putting fโ(๐ฅ)=0 3(๐ฅ^2โ4๐ฅ+3)=0 ๐ฅ^2โ4๐ฅ+3=0 ๐ฅ^2โ3๐ฅโ๐ฅ+3=0 ๐ (๐ฅ)=๐ฅ3 โ6๐ฅ2+9๐ฅ+15 Finding fโ(๐ฅ) fโ(๐ฅ)=๐(๐ฅ3 โ 6๐ฅ2+ 9๐ฅ + 15" " )/๐๐ฅ fโ(๐ฅ)=3๐ฅ^2โ12๐ฅ+9 fโ(๐ฅ)=3(๐ฅ^2โ4๐ฅ+3) Putting fโ(๐ฅ)=0 3(๐ฅ^2โ4๐ฅ+3)=0 ๐ฅ^2โ4๐ฅ+3=0 ๐ฅ^2โ3๐ฅโ๐ฅ+3=0 ๐ฅ(๐ฅโ3)โ1(๐ฅโ3)=0 (๐ฅโ1)(๐ฅโ3)=0 So, x = 1 & x = 3 Step 3: Finding fโโ(๐ฅ) fโ(๐ฅ)=3(๐ฅ^2โ4๐ฅ+3) fโโ(๐ฅ)=๐(3(๐ฅ^2โ4๐ฅ+3))/๐๐ฅ = 3 ๐(๐ฅ^2โ4๐ฅ+3)/๐๐ฅ = 3(2๐ฅโ4+0) = 6๐ฅโ12 โด fโโ(๐ฅ)=6๐ฅโ12 Putting ๐ฅ=1 in fโโ(๐ฅ) fโโ(1)=6(1)โ12 = 6 โ 12 = โ 6 < 0 Since fโโ(๐ฅ)<0 when ๐ฅ=1 โ ๐ฅ=1 is point of local maxima โด f(๐ฅ) is maximum at ๐=๐ Maximum value of f(๐ฅ) at ๐ฅ = 1 f(๐ฅ)=๐ฅ^3โ6๐ฅ^2+9๐ฅ+15 f(1)=(1)^3โ6(1)^2+9(1)+15 = 1 โ 6 + 9 + 15 = 19 Putting ๐ฅ=3 in fโโ(x) fโโ(๐ฅ)=6๐ฅโ12 fโโ(3)=6(3)โ12 = 18 โ 12 = 6 Since fโโ(๐ฅ)>0 when ๐ฅ=3 โ ๐ฅ=3 is point of local minima โด f(๐ฅ) is minimum at ๐=๐ Minimum value of f(๐ฅ) at ๐ฅ = 3 f(๐ฅ)=๐ฅ^3โ6๐ฅ^2+9๐ฅ+15 f(3)=(3)^3โ6(3)^2+9(3)+15 = 27 โ 54 + 27 + 15 = 15 Ex 6.5,3 (Method 1) 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 (๐ฅ) = ๐ฅ/2 + 2/๐ฅ , ๐ฅ > 0 g (๐ฅ) = ๐ฅ/2 + 2/๐ฅ , ๐ฅ > 0 Finding gโ(๐ฅ) gโ(๐ฅ)=๐/๐๐ฅ (๐ฅ/2+2/๐ฅ) =๐/๐๐ฅ (๐ฅ/2)+๐/๐๐ฅ (2๐ฅ^(โ1) ) =1/2โ2๐ฅ^(โ2) =1/2โ2/๐ฅ^2 Putting gโ(๐ฅ)=0 1/2โ2/๐ฅ^2 =0 (๐ฅ^2โ 4)/(2๐ฅ^2 )=0 ๐ฅ^2โ4=0 ร2๐ฅ^2 (๐ฅโ2)(๐ฅ+2)=0 So, ๐ฅ=2 & ๐ฅ=โ2 Since ๐ฅ>0 is given, we consider only ๐ฅ=2 Since, gโ(๐ฅ) changes sign from negative to positive โ ๐ฅ=2 is point of local minima โด g(๐) is minimum at ๐=๐ Minimum value of g(๐ฅ) is g(๐ฅ)=๐ฅ/2+2/๐ฅ Putting ๐ฅ=2 g(2)= 2/2+2/2 = 1 + 1 = 2 Ex 6.5,3 (Method 2) 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 (๐ฅ) = ๐ฅ/2 + 2/๐ฅ , ๐ฅ > 0 Ex 6.5,3 (Method 2) 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 (๐ฅ) = ๐ฅ/2 + 2/๐ฅ , ๐ฅ > 0 g (๐ฅ) = ๐ฅ/2 + 2/๐ฅ , ๐ฅ > 0 Finding gโ(๐ฅ) gโ(๐ฅ)=๐/๐๐ฅ (๐ฅ/2+2/๐ฅ) =๐/๐๐ฅ (๐ฅ/2)+๐/๐๐ฅ (2๐ฅ^(โ1) ) =1/2โ2๐ฅ^(โ2) =1/2โ2/๐ฅ^2 Putting gโ(๐ฅ)=0 1/2โ2/๐ฅ^2 =0 (๐ฅ^2โ 4)/(2๐ฅ^2 )=0 ๐ฅ^2โ4=0 ร2๐ฅ^2 (๐ฅโ2)(๐ฅ+2)=0 So, ๐ฅ=2 & ๐ฅ=โ2 Since ๐ฅ>0 is given, we consider only ๐ฅ=2 Finding gโโ(๐) gโ(๐ฅ)=1/2โ2/๐ฅ^2 gโโ(๐ฅ)=๐/๐๐ฅ (1/2โ2/๐ฅ^2 ) = 0 โ 2 . (โ2) ๐ฅ^(โ2โ1) = 4๐ฅ^(โ3) = 4/๐ฅ^3 Putting value ๐ฅ=2 in gโโ(๐ฅ) gโโ(๐ฅ)=4/(2)^3 = 4/8= 1/2 Since gโโ(๐ฅ)>0 when ๐ฅ=2 โ ๐ฅ=2 is Point of Local Minima โ g(๐ฅ) is Minimum at ๐=๐ Minimum value of g(๐ฅ) is g(๐ฅ)=๐ฅ/2+2/๐ฅ Putting ๐ฅ=2 g(2)= 2/2+2/2 = 1 + 1 = 2 Ex 6.5,3 (Method 1) 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 (๐ฅ) = 1/(๐ฅ^2 + 2) Step 1: Finding gโ(๐ฅ) gโ(๐ฅ)=๐/๐๐ฅ (1/(๐ฅ^2 + 2)) using quotient rule as (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃ โ ๐ฃ^โฒ ๐ข)/๐ฃ^2 gโฒ(๐ฅ)=(๐(1)/๐๐ฅ . (๐ฅ^2 + 2) โ ๐(๐ฅ^2 + 2)/๐๐ฅ . 1)/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( 0 . (๐ฅ^2 + 2) โ (2๐ฅ + 0))/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( 0 โ 2๐ฅ )/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( โ2๐ฅ )/(๐ฅ^2 + 2)^2 Step 2: Putting gโ(๐ฅ)=0 ( โ2๐ฅ )/(๐ฅ^2+2)^2 =0 โ2๐ฅ=0 ร(๐ฅ^2+2)^2 โ2๐ฅ=0 โ ๐ฅ=0 Step 3: โด ๐ฅ = 0 is point of local maxima g(๐ฅ) is maximum at ๐ = 0 Maximum value of g(๐ฅ) g(๐ฅ)=1/(๐ฅ^2 + 2) Putting ๐ฅ = 0 g(0)=1/(0^2 + 2) = 1/2 Maximum value is ๐/๐ Ex 6.5,3 (Method 2) 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 (๐ฅ) = 1/(๐ฅ^2 + 2) Step 1: Finding gโ(๐ฅ) gโ(๐ฅ)=๐/๐๐ฅ (1/(๐ฅ^2 + 2)) using quotient rule as (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃ โ ๐ฃ^โฒ ๐ข)/๐ฃ^2 gโฒ(๐ฅ)=(๐(1)/๐๐ฅ . (๐ฅ^2 + 2) โ ๐(๐ฅ^2 + 2)/๐๐ฅ . 1)/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( 0 . (๐ฅ^2 + 2) โ (2๐ฅ + 0))/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( 0 โ 2๐ฅ )/(๐ฅ^2 + 2)^2 gโฒ(๐ฅ)=( โ2๐ฅ )/(๐ฅ^2 + 2)^2 Step 2: Putting gโ(๐ฅ)=0 ( โ2๐ฅ )/(๐ฅ^2+2)^2 =0 โ2๐ฅ=0 ร(๐ฅ^2+2)^2 โ2๐ฅ=0 โ ๐ฅ=0 Step 3: Finding gโโ(๐ฅ) gโ(๐ฅ)=(โ2๐ฅ)/(๐ฅ^2+2)^2 using quotient rule as (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃโ๐ฃ^โฒ ๐ข)/๐ฃ^2 gโ(๐ฅ)=(๐(โ2๐ฅ)/๐๐ฅ . ใ (๐ฅ^2+2)ใ^2 โ (๐(๐ฅ^2+2)^2)/๐๐ฅ . (โ2๐ฅ))/((๐ฅ^2+2)^2 )^2 =(โ2 (๐ฅ^2+2)^2โ2 (๐ฅ^2+2)^(2โ1).๐(๐ฅ^2+2)/๐๐ฅ . (โ2๐ฅ))/((๐ฅ^2+2)^2 )^2 =(โ2 (๐ฅ^2+2)^2โ2 (๐ฅ^2+2)(2๐ฅ+0) (โ2๐ฅ))/(๐ฅ^2+2)^4 =(โ2 (๐ฅ^2+2)^2โ2 (๐ฅ^2+2)(2๐ฅ) (โ2๐ฅ))/(๐ฅ^2+2)^4 =(โ2 (๐ฅ^2+2)^2+8๐ฅ^2 (๐ฅ^2+2))/(๐ฅ^2+2)^4 =(โ2 (๐ฅ^2+2)[(๐ฅ^2+2)โ4๐ฅ^2 ])/(๐ฅ^2+2)^4 =(โ2 (๐ฅ^2+2)(โ3๐ฅ^2+2))/(๐ฅ^2+2)^4 =(โ2(โ3๐ฅ^2+2))/(๐ฅ^2+2)^3 Putting x = 0 in gโโ(x) gโโ(0)=(โ2(โ3(0) + 2))/(0^2+ 2)^3 =(โ2(0 + 2))/(2)^3 =(โ4)/8=(โ1)/2 Hence gโโ(๐ฅ)<0 when ๐ฅ = 0 โ ๐ฅ = 0 is point of local maxima โ g(๐ฅ) is maximum at ๐ฅ = 0 Maximum value of g(๐ฅ) g(๐ฅ)=1/(๐ฅ^2+2) Putting ๐ฅ = 0 g(0)=1/(0^2+2) = 1/2 Maximum value is ๐/๐ Ex 6.5,3 (Method 1) 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(๐ฅ) = ๐ฅโ(1โ๐ฅ), ๐ฅ > 0 Step 1: Finding fโ(๐ฅ) fโ(๐ฅ)=๐(๐ฅโ(1 โ ๐ฅ))/๐๐ฅ using product rule as (๐ข๐ฃ)^โฒ=๐ขโv+vโu fโ(๐ฅ)=๐(๐ฅ)/๐๐ฅ . โ(1โ๐ฅ) + ๐(โ(1 โ ๐ฅ))/๐๐ฅ . ๐ฅ = 1 . โ(1โ๐ฅ) + 1/(2โ(1 โ ๐ฅ)) . ๐(1 โ ๐ฅ)/๐๐ฅ . ๐ฅ = โ(1โ๐ฅ) + 1/(2โ(1 โ ๐ฅ)) (0 โ1) . ๐ฅ = โ(1โ๐ฅ) โ ๐ฅ/(2โ(1 โ ๐ฅ)) = (2(โ(1 โ ๐ฅ) )^2โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2(1 โ ๐ฅ) โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2 โ 2๐ฅ โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ)) Step 2: Putting fโ(๐ฅ)=0 (2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ))=0 2 โ 3๐ฅ = 0 ร 2โ(1โ๐ฅ) 2 โ 3๐ฅ=0 โ 3๐ฅ=โ2 ๐ฅ =2/3 Step 3: โ ๐ฅ = 2/3 is point of local maxima โ f(๐ฅ) is maximum at ๐ฅ = 2/3 Step 4: Finding Maximum value of f(๐ฅ)=๐ฅโ(1โ๐ฅ) Putting ๐ฅ=2/3 f(2/3)=2/3 โ(1โ2/3) =2/3 โ((3โ2)/3) =2/3 โ(1/3) =2/(3โ3) =2/(3โ3) ร โ3/โ3 =(2โ3)/9 Maximum value of f(๐) is (๐โ๐)/๐ at x = ๐/๐ Ex 6.5,3 (Method 2) 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(๐ฅ) = ๐ฅโ(1โ๐ฅ), ๐ฅ > 0 Step 1: Finding fโ(๐ฅ) fโ(๐ฅ)=๐(๐ฅโ(1 โ ๐ฅ))/๐๐ฅ using product rule as (๐ข๐ฃ)^โฒ=๐ขโv+vโu fโ(๐ฅ)=๐(๐ฅ)/๐๐ฅ . โ(1โ๐ฅ) + ๐(โ(1 โ ๐ฅ))/๐๐ฅ . ๐ฅ = 1 . โ(1โ๐ฅ) + 1/(2โ(1 โ ๐ฅ)) . ๐(1 โ ๐ฅ)/๐๐ฅ . ๐ฅ = โ(1โ๐ฅ) + 1/(2โ(1 โ ๐ฅ)) (0 โ1) . ๐ฅ = โ(1โ๐ฅ) โ ๐ฅ/(2โ(1 โ ๐ฅ)) = (2(โ(1 โ ๐ฅ) )^2โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2(1 โ ๐ฅ) โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2 โ 2๐ฅ โ ๐ฅ)/(2โ(1 โ ๐ฅ)) = (2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ)) Step 2: Putting fโ(๐ฅ)=0 (2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ))=0 2 โ 3๐ฅ = 0 ร 2โ(1โ๐ฅ) 2 โ 3๐ฅ=0 โ 3๐ฅ=โ2 ๐ฅ =2/3 Step 3: Finding fโโ(๐ฅ) fโ(๐ฅ)=(2โ3๐ฅ)/(2โ(1โ๐ฅ)) fโโ(๐ฅ)=๐/๐๐ฅ ((2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ))) using quotient rule as (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃ โ๐ฃ^โฒ ๐ข)/๐ฃ^2 = 1/2 [(๐(2 โ 3๐ฅ)/๐๐ฅ . โ(1 โ ๐ฅ) โ ๐(โ(1 โ ๐ฅ))/๐๐ฅ . (2 โ 3๐ฅ))/(โ(1 โ ๐ฅ))^2 ] = 1/2 [((0 โ 3) โ(1 โ ๐ฅ) โ 1/(2โ(1 โ ๐ฅ)) . ๐(1 โ ๐ฅ)/๐๐ฅ . (2 โ 3๐ฅ))/((1 โ ๐ฅ) )] = 1/2 [(โ3โ(1 โ ๐ฅ) โ 1/(2โ(1 โ ๐ฅ)) (0 โ 1) . (2 โ 3๐ฅ))/((1 โ ๐ฅ) )] = 1/2 [(โ3โ(1 โ ๐ฅ) + (2 โ 3๐ฅ)/(2โ(1 โ ๐ฅ)) )/(1 โ ๐ฅ)] = 1/2 [((โ3โ(1 โ ๐ฅ)) (2โ(1 โ ๐ฅ)) + 2 โ 3๐ฅ )/(2(1 โ ๐ฅ) โ(1 โ ๐ฅ))] = 1/2 [(โ6(1 โ ๐ฅ) + 2 โ 3๐ฅ )/(2(1 โ ๐ฅ) โ(1 โ ๐ฅ))] = 1/2 [(โ6 + 6๐ฅ + 2 โ 3๐ฅ )/(2(1 โ ๐ฅ) โ(1 โ ๐ฅ))] = 1/4 [(โ4 + 3๐ฅ )/(1 + ๐ฅ)^(3/2) ] Hence fโโ(๐ฅ)=1/4 [(โ4 + 3๐ฅ )/(1 + ๐ฅ)^(3/2) ] Putting ๐ฅ=2/3 fโโ(2/3)=1/4 [(โ4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(โ4 + 2)/(5/3)^(3/2) ] =1/4 [(โ2)/(5/3)^(3/2) ] = (โ 1)/2 (3/5)^(3/2) < 0 โด fโโ(๐ฅ)<0 when ๐ฅ = 2/3 Hence, ๐ฅ=2/3 is the maxima Step 4: Finding Maximum value of f(๐ฅ)=๐ฅโ(1โ๐ฅ) Putting ๐ฅ=2/3 f(2/3)=2/3 โ(1โ2/3) =2/3 โ((3 โ 2)/3) =2/3 โ(1/3) =2/(3โ3) =2/(3โ3) ร โ3/โ3 =(2โ3)/9 Maximum value of f(๐) is (๐โ๐)/๐ at x = ๐/๐

Ex 6.5