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

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  1. Chapter 6 Class 12 Application of Derivatives
  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 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 โˆˆ 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 โˆˆ 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 โˆˆ 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^3โˆ’3๐‘ฅ)/๐‘‘๐‘ฅ gโ€™(๐‘ฅ)=3๐‘ฅ^2โˆ’3 Step 2: Putting gโ€™(๐‘ฅ)=0 3๐‘ฅ^2โˆ’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 (โˆ’)๐‘ฃ๐‘’ when ๐‘ฅ=โˆ’1 โ‡’ ๐‘ฅ=โˆ’1 is point of local maxima โ‡’ g(๐‘ฅ) is maximum at ๐‘ฅ=โˆ’1 Local minimum value of g(๐‘ฅ)=๐‘ฅ^3โˆ’3๐‘ฅ at x = 1 g(1)=(1)^3โˆ’3(1)=1โˆ’3=โˆ’๐Ÿ Local maximum value of g(๐‘ฅ)=๐‘ฅ^3โˆ’3๐‘ฅ at x = โ€“1 g(โˆ’1)= (โˆ’1)^3โˆ’3(โˆ’1)=โˆ’1+3 =๐Ÿ 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^3โˆ’3๐‘ฅ)/๐‘‘๐‘ฅ gโ€™(๐‘ฅ)=3๐‘ฅ^2โˆ’3 Step 2: Putting gโ€™(๐‘ฅ)=0 3๐‘ฅ^2โˆ’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^2โˆ’3 gโ€™โ€™(๐‘ฅ)=๐‘‘(3๐‘ฅ^2โˆ’3)/๐‘‘๐‘ฅ = 6๐‘ฅโˆ’0 = 6๐‘ฅ Putting ๐’™=๐Ÿ 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^3โˆ’3๐‘ฅ g(1)=(1)^3โˆ’3(1) =1โˆ’3 =โˆ’๐Ÿ Putting ๐’™=โˆ’๐Ÿ in gโ€™โ€™(x) gโ€™โ€™(โˆ’1)=6(โˆ’1)= โ€“6 < 0 Thus, gโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ=โˆ’1 โ‡’ ๐‘ฅ=โˆ’1 is point of local maxima & g(๐‘ฅ) is maximum at ๐‘ฅ=โˆ’1 Local minimum value g(๐‘ฅ)=๐‘ฅ^3โˆ’3๐‘ฅ g(โˆ’1)=(โˆ’1)^3โˆ’3(โˆ’1) =โˆ’1+3 =๐Ÿ 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„Ž(๐‘ฅ)=sinโก๐‘ฅ+cosโก๐‘ฅ, 0<๐‘ฅ<๐œ‹/2 โ„Ž(๐‘ฅ)=sinโก๐‘ฅ+cosโก๐‘ฅ, 0<๐‘ฅ<๐œ‹/2 Step 1: Finding โ„Žโ€ฒ(๐‘ฅ) โ„Žโ€ฒ(๐‘ฅ)=๐‘‘(sinโก๐‘ฅ+ cosโก๐‘ฅ" " )/๐‘‘๐‘ฅ โ„Ž^โ€ฒ (๐‘ฅ)=cosโก๐‘ฅโˆ’sinโก๐‘ฅ Step 2: Putting โ„Ž(๐‘ฅ)=0 cosโก๐‘ฅโˆ’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 โˆด ๐‘ฅ=45ยฐ= ๐œ‹/4 Step 3: Since sign of fโ€™(x) changes from positive to negative, x = ๐œ‹/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‘œ๐‘ ๐‘ฅ f(ฯ€/4)=๐‘ ๐‘–๐‘›(ฯ€/4)+๐‘๐‘œ๐‘ (ฯ€/4) = 1/โˆš2+1/โˆš2 = 2/โˆš2 = โˆš๐Ÿ 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„Ž(๐‘ฅ)=sinโก๐‘ฅ+cosโก๐‘ฅ, 0<๐‘ฅ<๐œ‹/2 โ„Ž(๐‘ฅ)=sinโก๐‘ฅ+cosโก๐‘ฅ, 0<๐‘ฅ<๐œ‹/2 Step 1: Finding โ„Žโ€ฒ(๐‘ฅ) โ„Žโ€ฒ(๐‘ฅ)=๐‘‘(sinโก๐‘ฅ+ cosโก๐‘ฅ" " )/๐‘‘๐‘ฅ โ„Ž^โ€ฒ (๐‘ฅ)=cosโก๐‘ฅโˆ’sinโก๐‘ฅ Step 2: Putting โ„Ž(๐‘ฅ)=0 cosโก๐‘ฅโˆ’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 โˆด ๐‘ฅ=45ยฐ= ๐œ‹/4 Step 3: Finding hโ€™โ€™(๐‘ฅ) hโ€™(๐‘ฅ)=๐‘๐‘œ๐‘ ๐‘ฅโˆ’๐‘ ๐‘–๐‘›๐‘ฅ hโ€™โ€™(๐‘ฅ)=โˆ’๐‘ ๐‘–๐‘›๐‘ฅโˆ’๐‘๐‘œ๐‘ ๐‘ฅ Putting ๐‘ฅ=ฯ€/4 hโ€™โ€™(ฯ€/4)=โˆ’sin(ฯ€/4)โˆ’๐‘๐‘œ๐‘ (ฯ€/4) = โ€“ 1/โˆš2โˆ’1/โˆš2 = (โˆ’2)/โˆš2 = โ€“ โˆš2 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Ÿ’ f(๐‘ฅ)=๐‘ ๐‘–๐‘›๐‘ฅ+๐‘๐‘œ๐‘ ๐‘ฅ f(ฯ€/4)=๐‘ ๐‘–๐‘›(ฯ€/4)+๐‘๐‘œ๐‘ (ฯ€/4) = 1/โˆš2+1/โˆš2 = 2/โˆš2 = โˆš๐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: (iv) f (๐‘ฅ)=sinโก๐‘ฅ โ€“cosโก๐‘ฅ, 0<๐‘ฅ<2 ๐œ‹ f (๐‘ฅ)=sinโก๐‘ฅ โ€“cosโก๐‘ฅ, 0<๐‘ฅ<2 ๐œ‹ Finding fโ€™(๐‘ฅ) fโ€™(๐‘ฅ)=cosโก๐‘ฅโˆ’(โˆ’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=โˆ’sinโก๐‘ฅ 1=(โˆ’sinโก๐‘ฅ)/cosโก๐‘ฅ (โˆ’sinโก๐‘ฅ)/cosโก๐‘ฅ =1 โ€“ tan ๐‘ฅ=1 tan ๐‘ฅ=โˆ’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œ‹/4 ๐‘œ๐‘Ÿ 7๐œ‹/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)=โˆ’sinโก๐‘ฅ+cosโก๐‘ฅ Putting ๐’™ = ๐Ÿ‘๐…/๐Ÿ’ fโ€™โ€™(3๐œ‹/4)=โˆ’๐‘ ๐‘–๐‘›(3๐œ‹/4)+๐‘๐‘œ๐‘ (3๐œ‹/4) =โˆ’๐‘ ๐‘–๐‘›(๐œ‹โˆ’๐œ‹/4)+๐‘๐‘œ๐‘ (๐œ‹โˆ’ ๐œ‹/4) =โˆ’๐‘ ๐‘–๐‘›(๐œ‹/4)+(โˆ’๐‘๐‘œ๐‘  ๐œ‹/4) =(โˆ’1)/โˆš2โˆ’1/โˆš2 =(โˆ’2)/โˆš2 =โˆ’โˆš2 < 0 Hence fโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ = 3๐œ‹/4 Thus ๐‘ฅ = 3๐œ‹/4 is point of local maxima โˆด f(๐‘ฅ) is maximum value at ๐‘ฅ = 3๐œ‹/4 The local maximum value is f(๐‘ฅ)=sinโก๐‘ฅโˆ’cosโก๐‘ฅ f(3๐œ‹/4)=๐‘ ๐‘–๐‘›(3๐œ‹/4)โˆ’๐‘๐‘œ๐‘ (3๐œ‹/4) =๐‘ ๐‘–๐‘›(๐œ‹โˆ’๐œ‹/4)โˆ’๐‘๐‘œ๐‘ (๐œ‹โˆ’๐œ‹/4) =๐‘ ๐‘–๐‘›(๐œ‹/4)โˆ’(โˆ’๐‘๐‘œ๐‘  ๐œ‹/4) =๐‘ ๐‘–๐‘› ๐œ‹/4+๐‘๐‘œ๐‘  ๐œ‹/4 =1/โˆš2+1/โˆš2 =2/โˆš2 =โˆš2 Now, for ๐’™ = ๐Ÿ•๐…/๐Ÿ’ fโ€™โ€™(๐‘ฅ)=โˆ’sinโก๐‘ฅ+cosโก๐‘ฅ Putting ๐‘ฅ = 7๐œ‹/4 fโ€™โ€™(7๐œ‹/4)=โˆ’sinโก(7๐œ‹/4)+cosโก(7๐œ‹/4) fโ€™โ€™(7๐œ‹/4)=โˆ’๐‘ ๐‘–๐‘›(2๐œ‹โˆ’๐œ‹/4)+๐‘๐‘œ๐‘ (2๐œ‹โˆ’๐œ‹/4) =โˆ’(โˆ’๐‘ ๐‘–๐‘›(๐œ‹/4))+๐‘๐‘œ๐‘ (๐œ‹/4) =๐‘ ๐‘–๐‘› ๐œ‹/4+๐‘๐‘œ๐‘  ๐œ‹/4 =1/โˆš2 + 1/โˆš2 =2/โˆš2 =โˆš2 > 0 fโ€™โ€™(๐‘ฅ)>0 when ๐‘ฅ = 7๐œ‹/4 Thus ๐‘ฅ = 7๐œ‹/4 is point of local minima f(๐‘ฅ) has minimum value at ๐‘ฅ = 7๐œ‹/4 Local minimum value is f(๐‘ฅ)=๐‘ ๐‘–๐‘›๐‘ฅโˆ’๐‘๐‘œ๐‘ ๐‘ฅ f(7๐œ‹/4)=๐‘ ๐‘–๐‘›(7๐œ‹/4)โˆ’๐‘๐‘œ๐‘ (7๐œ‹/4) =๐‘ ๐‘–๐‘›(2๐œ‹โˆ’๐œ‹/4)โˆ’๐‘๐‘œ๐‘ (2๐œ‹โˆ’๐œ‹/4) =โˆ’๐‘ ๐‘–๐‘›(๐œ‹/4)โˆ’๐‘๐‘œ๐‘ (๐œ‹/4) =(โˆ’1)/โˆš2 โˆ’ 1/โˆš2 =(โˆ’2)/โˆš2 =โˆ’โˆš2 Thus, f(๐‘ฅ) is maximum at x = ๐Ÿ‘๐…/๐Ÿ’ and maximum value is โˆš๐Ÿ & f(๐‘ฅ) is minimum at x = ๐Ÿ•๐…/๐Ÿ’ and maximum value is โ€“โˆš๐Ÿ 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^2โˆ’12๐‘ฅ+9 fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) Putting fโ€™(๐‘ฅ)=0 3(๐‘ฅ^2โˆ’4๐‘ฅ+3)=0 ๐‘ฅ^2โˆ’4๐‘ฅ+3=0 ๐‘ฅ^2โˆ’3๐‘ฅโˆ’๐‘ฅ+3=0 ๐‘ฅ(๐‘ฅโˆ’3)โˆ’1(๐‘ฅโˆ’3)=0 (๐‘ฅโˆ’1)(๐‘ฅโˆ’3)=0 So, x = 1 & x = 3 Step 3: Using first derivative test at x = 1 & x = โˆ’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^3โˆ’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^3โˆ’6๐‘ฅ^2+9๐‘ฅ+15 f(๐‘ฅ)=(3)^3โˆ’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^2โˆ’12๐‘ฅ+9 fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) Putting fโ€™(๐‘ฅ)=0 3(๐‘ฅ^2โˆ’4๐‘ฅ+3)=0 ๐‘ฅ^2โˆ’4๐‘ฅ+3=0 ๐‘ฅ^2โˆ’3๐‘ฅโˆ’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^2โˆ’12๐‘ฅ+9 fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) Putting fโ€™(๐‘ฅ)=0 3(๐‘ฅ^2โˆ’4๐‘ฅ+3)=0 ๐‘ฅ^2โˆ’4๐‘ฅ+3=0 ๐‘ฅ^2โˆ’3๐‘ฅโˆ’๐‘ฅ+3=0 ๐‘ฅ(๐‘ฅโˆ’3)โˆ’1(๐‘ฅโˆ’3)=0 (๐‘ฅโˆ’1)(๐‘ฅโˆ’3)=0 So, x = 1 & x = 3 Step 3: Finding fโ€™โ€™(๐‘ฅ) fโ€™(๐‘ฅ)=3(๐‘ฅ^2โˆ’4๐‘ฅ+3) fโ€™โ€™(๐‘ฅ)=๐‘‘(3(๐‘ฅ^2โˆ’4๐‘ฅ+3))/๐‘‘๐‘ฅ = 3 ๐‘‘(๐‘ฅ^2โˆ’4๐‘ฅ+3)/๐‘‘๐‘ฅ = 3(2๐‘ฅโˆ’4+0) = 6๐‘ฅโˆ’12 โˆด fโ€™โ€™(๐‘ฅ)=6๐‘ฅโˆ’12 Putting ๐‘ฅ=1 in fโ€™โ€™(๐‘ฅ) fโ€™โ€™(1)=6(1)โˆ’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 โˆด f(๐‘ฅ) is maximum at ๐’™=๐Ÿ Maximum value of f(๐‘ฅ) at ๐‘ฅ = 1 f(๐‘ฅ)=๐‘ฅ^3โˆ’6๐‘ฅ^2+9๐‘ฅ+15 f(1)=(1)^3โˆ’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‘ฅโˆ’12 fโ€™โ€™(3)=6(3)โˆ’12 = 18 โ€“ 12 = 6 Since fโ€™โ€™(๐‘ฅ)>0 when ๐‘ฅ=3 โ‡’ ๐‘ฅ=3 is point of local minima โˆด f(๐‘ฅ) is minimum at ๐’™=๐Ÿ‘ Minimum value of f(๐‘ฅ) at ๐‘ฅ = 3 f(๐‘ฅ)=๐‘ฅ^3โˆ’6๐‘ฅ^2+9๐‘ฅ+15 f(3)=(3)^3โˆ’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^(โˆ’1) ) =1/2โˆ’2๐‘ฅ^(โˆ’2) =1/2โˆ’2/๐‘ฅ^2 Putting gโ€™(๐‘ฅ)=0 1/2โˆ’2/๐‘ฅ^2 =0 (๐‘ฅ^2โˆ’ 4)/(2๐‘ฅ^2 )=0 ๐‘ฅ^2โˆ’4=0 ร—2๐‘ฅ^2 (๐‘ฅโˆ’2)(๐‘ฅ+2)=0 So, ๐‘ฅ=2 & ๐‘ฅ=โˆ’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 โˆด g(๐’™) is minimum at ๐’™=๐Ÿ 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^(โˆ’1) ) =1/2โˆ’2๐‘ฅ^(โˆ’2) =1/2โˆ’2/๐‘ฅ^2 Putting gโ€™(๐‘ฅ)=0 1/2โˆ’2/๐‘ฅ^2 =0 (๐‘ฅ^2โˆ’ 4)/(2๐‘ฅ^2 )=0 ๐‘ฅ^2โˆ’4=0 ร—2๐‘ฅ^2 (๐‘ฅโˆ’2)(๐‘ฅ+2)=0 So, ๐‘ฅ=2 & ๐‘ฅ=โˆ’2 Since ๐‘ฅ>0 is given, we consider only ๐‘ฅ=2 Finding gโ€™โ€™(๐’™) gโ€™(๐‘ฅ)=1/2โˆ’2/๐‘ฅ^2 gโ€™โ€™(๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ (1/2โˆ’2/๐‘ฅ^2 ) = 0 โ€“ 2 . (โˆ’2) ๐‘ฅ^(โˆ’2โˆ’1) = 4๐‘ฅ^(โˆ’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Ÿ 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 โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 gโ€ฒ(๐‘ฅ)=(๐‘‘(1)/๐‘‘๐‘ฅ . (๐‘ฅ^2 + 2) โˆ’ ๐‘‘(๐‘ฅ^2 + 2)/๐‘‘๐‘ฅ . 1)/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( 0 . (๐‘ฅ^2 + 2) โˆ’ (2๐‘ฅ + 0))/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( 0 โˆ’ 2๐‘ฅ )/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( โˆ’2๐‘ฅ )/(๐‘ฅ^2 + 2)^2 Step 2: Putting gโ€™(๐‘ฅ)=0 ( โˆ’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: โˆด 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Ÿ/๐Ÿ 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 โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 gโ€ฒ(๐‘ฅ)=(๐‘‘(1)/๐‘‘๐‘ฅ . (๐‘ฅ^2 + 2) โˆ’ ๐‘‘(๐‘ฅ^2 + 2)/๐‘‘๐‘ฅ . 1)/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( 0 . (๐‘ฅ^2 + 2) โˆ’ (2๐‘ฅ + 0))/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( 0 โˆ’ 2๐‘ฅ )/(๐‘ฅ^2 + 2)^2 gโ€ฒ(๐‘ฅ)=( โˆ’2๐‘ฅ )/(๐‘ฅ^2 + 2)^2 Step 2: Putting gโ€™(๐‘ฅ)=0 ( โˆ’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)=(โˆ’2๐‘ฅ)/(๐‘ฅ^2+2)^2 using quotient rule as (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃโˆ’๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 gโ€™(๐‘ฅ)=(๐‘‘(โˆ’2๐‘ฅ)/๐‘‘๐‘ฅ . ใ€– (๐‘ฅ^2+2)ใ€—^2 โˆ’ (๐‘‘(๐‘ฅ^2+2)^2)/๐‘‘๐‘ฅ . (โˆ’2๐‘ฅ))/((๐‘ฅ^2+2)^2 )^2 =(โˆ’2 (๐‘ฅ^2+2)^2โˆ’2 (๐‘ฅ^2+2)^(2โˆ’1).๐‘‘(๐‘ฅ^2+2)/๐‘‘๐‘ฅ . (โˆ’2๐‘ฅ))/((๐‘ฅ^2+2)^2 )^2 =(โˆ’2 (๐‘ฅ^2+2)^2โˆ’2 (๐‘ฅ^2+2)(2๐‘ฅ+0) (โˆ’2๐‘ฅ))/(๐‘ฅ^2+2)^4 =(โˆ’2 (๐‘ฅ^2+2)^2โˆ’2 (๐‘ฅ^2+2)(2๐‘ฅ) (โˆ’2๐‘ฅ))/(๐‘ฅ^2+2)^4 =(โˆ’2 (๐‘ฅ^2+2)^2+8๐‘ฅ^2 (๐‘ฅ^2+2))/(๐‘ฅ^2+2)^4 =(โˆ’2 (๐‘ฅ^2+2)[(๐‘ฅ^2+2)โˆ’4๐‘ฅ^2 ])/(๐‘ฅ^2+2)^4 =(โˆ’2 (๐‘ฅ^2+2)(โˆ’3๐‘ฅ^2+2))/(๐‘ฅ^2+2)^4 =(โˆ’2(โˆ’3๐‘ฅ^2+2))/(๐‘ฅ^2+2)^3 Putting x = 0 in gโ€™โ€™(x) gโ€™โ€™(0)=(โˆ’2(โˆ’3(0) + 2))/(0^2+ 2)^3 =(โˆ’2(0 + 2))/(2)^3 =(โˆ’4)/8=(โˆ’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Ÿ/๐Ÿ 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‘ฅโˆš(1โˆ’๐‘ฅ), ๐‘ฅ > 0 Step 1: Finding fโ€™(๐‘ฅ) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅโˆš(1 โˆ’ 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 . โˆš(1โˆ’๐‘ฅ) + ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฅ = 1 . โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) . ๐‘‘(1 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) (0 โˆ’1) . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) โ€“ ๐‘ฅ/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(โˆš(1 โˆ’ ๐‘ฅ) )^2โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(1 โˆ’ ๐‘ฅ) โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 2๐‘ฅ โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) Step 2: Putting fโ€™(๐‘ฅ)=0 (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ))=0 2 โ€“ 3๐‘ฅ = 0 ร— 2โˆš(1โˆ’๐‘ฅ) 2 โ€“ 3๐‘ฅ=0 โ€“ 3๐‘ฅ=โˆ’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‘ฅโˆš(1โˆ’๐‘ฅ) Putting ๐‘ฅ=2/3 f(2/3)=2/3 โˆš(1โˆ’2/3) =2/3 โˆš((3โˆ’2)/3) =2/3 โˆš(1/3) =2/(3โˆš3) =2/(3โˆš3) ร— โˆš3/โˆš3 =(2โˆš3)/9 Maximum value of f(๐’™) is (๐Ÿโˆš๐Ÿ‘)/๐Ÿ— at x = ๐Ÿ/๐Ÿ‘ 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‘ฅโˆš(1โˆ’๐‘ฅ), ๐‘ฅ > 0 Step 1: Finding fโ€™(๐‘ฅ) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅโˆš(1 โˆ’ 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 . โˆš(1โˆ’๐‘ฅ) + ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฅ = 1 . โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) . ๐‘‘(1 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) (0 โˆ’1) . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) โ€“ ๐‘ฅ/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(โˆš(1 โˆ’ ๐‘ฅ) )^2โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(1 โˆ’ ๐‘ฅ) โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 2๐‘ฅ โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) Step 2: Putting fโ€™(๐‘ฅ)=0 (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ))=0 2 โ€“ 3๐‘ฅ = 0 ร— 2โˆš(1โˆ’๐‘ฅ) 2 โ€“ 3๐‘ฅ=0 โ€“ 3๐‘ฅ=โˆ’2 ๐‘ฅ =2/3 Step 3: Finding fโ€™โ€™(๐‘ฅ) fโ€™(๐‘ฅ)=(2โˆ’3๐‘ฅ)/(2โˆš(1โˆ’๐‘ฅ)) fโ€™โ€™(๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ ((2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ))) using quotient rule as (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃ โˆ’๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 = 1/2 [(๐‘‘(2 โˆ’ 3๐‘ฅ)/๐‘‘๐‘ฅ . โˆš(1 โˆ’ ๐‘ฅ) โˆ’ ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ . (2 โˆ’ 3๐‘ฅ))/(โˆš(1 โˆ’ ๐‘ฅ))^2 ] = 1/2 [((0 โˆ’ 3) โˆš(1 โˆ’ ๐‘ฅ) โˆ’ 1/(2โˆš(1 โˆ’ ๐‘ฅ)) . ๐‘‘(1 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . (2 โˆ’ 3๐‘ฅ))/((1 โˆ’ ๐‘ฅ) )] = 1/2 [(โˆ’3โˆš(1 โˆ’ ๐‘ฅ) โˆ’ 1/(2โˆš(1 โˆ’ ๐‘ฅ)) (0 โˆ’ 1) . (2 โˆ’ 3๐‘ฅ))/((1 โˆ’ ๐‘ฅ) )] = 1/2 [(โˆ’3โˆš(1 โˆ’ ๐‘ฅ) + (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) )/(1 โˆ’ ๐‘ฅ)] = 1/2 [((โˆ’3โˆš(1 โˆ’ ๐‘ฅ)) (2โˆš(1 โˆ’ ๐‘ฅ)) + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/2 [(โˆ’6(1 โˆ’ ๐‘ฅ) + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/2 [(โˆ’6 + 6๐‘ฅ + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/4 [(โˆ’4 + 3๐‘ฅ )/(1 + ๐‘ฅ)^(3/2) ] Hence fโ€™โ€™(๐‘ฅ)=1/4 [(โˆ’4 + 3๐‘ฅ )/(1 + ๐‘ฅ)^(3/2) ] Putting ๐‘ฅ=2/3 fโ€™โ€™(2/3)=1/4 [(โˆ’4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(โˆ’4 + 2)/(5/3)^(3/2) ] =1/4 [(โˆ’2)/(5/3)^(3/2) ] = (โˆ’ 1)/2 (3/5)^(3/2) < 0 โˆด 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‘ฅโˆš(1โˆ’๐‘ฅ) Putting ๐‘ฅ=2/3 f(2/3)=2/3 โˆš(1โˆ’2/3) =2/3 โˆš((3 โˆ’ 2)/3) =2/3 โˆš(1/3) =2/(3โˆš3) =2/(3โˆš3) ร— โˆš3/โˆš3 =(2โˆš3)/9 Maximum value of f(๐’™) is (๐Ÿโˆš๐Ÿ‘)/๐Ÿ— at x = ๐Ÿ/๐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 9 years. He provides courses for Maths and Science at Teachoo.