1. Class 12
2. Important Question for exams Class 12
3. Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.5,7 Find both the maximum value and the minimum value of 3๐ฅ4 โ 8๐ฅ3 + 12๐ฅ2 โ 48๐ฅ + 25 on the interval [0, 3]. Let f(x) = 3๐ฅ4 โ 8๐ฅ3 + 12๐ฅ2 โ 48๐ฅ + 25 ๐ฅ โ [0, 3] Step 1: Finding fโ(๐ฅ) fโ(๐ฅ)=๐(3๐ฅ^4 โ 8๐ฅ^3+ 12๐ฅ^2โ 48๐ฅ + 25)/๐๐ฅ fโ(๐ฅ)=3 ร4๐ฅ^3โ8 ร3๐ฅ^2+12 ร2๐ฅโ48+0 fโ(๐ฅ)=12๐ฅ^3โ24๐ฅ^2+24๐ฅโ48 fโ(๐ฅ)=12(๐ฅ^3โ2๐ฅ^2+2๐ฅโ4) Step 2: Putting fโ(๐ฅ)=0 12(๐ฅ^3โ2๐ฅ^2+2๐ฅโ4)=0 ๐ฅ^3โ2๐ฅ^2+2๐ฅโ4=0 ๐ฅ^2 (๐ฅโ2)+2(๐ฅโ2)=0 (๐ฅ^2+2)(๐ฅโ2)=0 Since ๐ฅ^2=โ2 is not possible Thus ๐ฅ=2 is only critical point Step 3: Since are given interval ๐ฅ โ [0 , 3] Hence , calculating f(๐ฅ) at ๐ฅ = 0 , 2 & 3 Hence, Minimum value of f(๐ฅ) is โ39 at ๐ = 2 Maximum value of f(๐ฅ) is 25 at ๐ = 0

Chapter 6 Class 12 Application of Derivatives

Class 12
Important Question for exams Class 12