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, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum & the local minimum values, as the case may be: (i) f (๐ฅ)=๐ฅ2 f(๐ฅ)=๐ฅ^2 Finding fโ(x) fโ(x) = 2x Putting fโ(x) = 0 2x = 0 x = 0 Finding fโโ(x) fโ(x) = 2x Differentiating again fโโ(x) = 2 Since fโโ(x) > 0 for x = 0 So, f(x) is minimum at x = 0 Minimum value of f(x) at x = 0 f(x) = x2 f(0) = 02 = 0 So, Minimum value of f(x) = 0 There is no maximum value 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: (ii) ๐(๐ฅ)=๐ฅ3 โ3๐ฅ ๐(๐ฅ)=๐ฅ3 โ3๐ฅ Finding gโ(๐) gโ(๐ฅ)=๐(๐ฅ^3 โ 3๐ฅ)/๐๐ฅ gโ(๐ฅ)=3๐ฅ^2โ3 Putting gโ(๐)=๐ 3๐ฅ^2โ3=0 3๐ฅ^2=3 ๐ฅ^2=3/3 ๐ฅ^2=1 ๐ฅ=ยฑ1 So, x = 1 & x = โ1 Finding gโโ(๐) gโ(๐ฅ)=3๐ฅ^2โ3 gโโ(๐ฅ)=๐(3๐ฅ^2โ3)/๐๐ฅ = 6๐ฅโ0 = 6๐ฅ Putting ๐=๐ in gโโ(x) gโโ(1)=6(1)= 6 > 0 Thus, gโโ(๐ฅ)>0 when ๐ฅ=1 โ ๐ฅ=1 is point of local minima & g(๐ฅ) is minimum at ๐ฅ=1 Local minimum value g(๐ฅ)=๐ฅ^3โ3๐ฅ g(1)=(1)^3โ3(1) =1โ3 =โ๐ Putting ๐=โ๐ in gโโ(x) gโโ(โ1)=6(โ1)= โ6 < 0 Thus, gโโ(๐ฅ)<0 when ๐ฅ=โ1 โ ๐ฅ=โ1 is point of local maxima & g(๐ฅ) is maximum at ๐ฅ=โ1 Local minimum value g(๐ฅ)=๐ฅ^3โ3๐ฅ g(โ1)=(โ1)^3โ3(โ1) =โ1+3 =๐
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