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, 7 Find both the maximum value and the minimum value of 3๐ฅ4 โ 8๐ฅ3 + 12๐ฅ2 โ 48๐ฅ + 25 on the interval [0, 3]. Let f(x) = 3๐ฅ4 โ 8๐ฅ3 + 12๐ฅ2 โ 48๐ฅ + 25, where ๐ฅ โ [0, 3] Finding fโ(๐) fโ(๐ฅ)=๐(3๐ฅ^4 โ 8๐ฅ^3 + 12๐ฅ^2 โ 48๐ฅ + 25)/๐๐ฅ fโ(๐ฅ)=3 ร4๐ฅ^3โ8 ร3๐ฅ^2+12 ร2๐ฅโ48+0 fโ(๐ฅ)=12๐ฅ^3โ24๐ฅ^2+24๐ฅโ48 fโ(๐ฅ)=12(๐ฅ^3โ2๐ฅ^2+2๐ฅโ4) Putting fโ(๐)=๐ 12(๐ฅ^3โ2๐ฅ^2+2๐ฅโ4)=0 ๐ฅ^3โ2๐ฅ^2+2๐ฅโ4=0 ๐ฅ^2 (๐ฅโ2)+2(๐ฅโ2)=0 (๐ฅ^2+2)(๐ฅโ2)=0 Since ๐ฅ^2=โ2 is not possible Thus ๐ฅ=2 is only critical point v๐ฅ^2+2=0 ๐ฅ^2=โ2 ๐ฅโ2=0 ๐ฅ=2 Square of number can not be negative Since are given interval ๐ฅ โ [0 , 3] Hence , calculating f(๐ฅ) at ๐ฅ = 0 , 2 & 3 Hence, Minimum value of f(๐ฅ) is โ39 at ๐ = 2 Maximum value of f(๐ฅ) is 25 at ๐ = 0
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