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

Transcript

Misc 13 Find the points at which the function f given by f (๐ฅ) = (๐ฅโ2)^4 (๐ฅ+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(๐ฅ)= (๐ฅโ2)^4 (๐ฅ+1)3 Step 1: Finding fโ(๐ฅ) fโ(๐ฅ) = (๐ ((๐ฅโ2)^4 (๐ฅ+1)^3 ))/๐๐ฅ using product rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข = ใ((๐ฅโ2)^4 )^โฒ (๐ฅ+1)ใ^3+((๐ฅ+1)^3 )^โฒ (๐ฅโ2)^4 = 4(๐ฅโ2)^3 (๐ฅ+1)^3+3(๐ฅ+1)^2 (๐ฅโ2)^4 = (๐ฅโ2)^3 (๐ฅ+1)^2 [4(๐ฅ+1)+3(๐ฅโ2)] = (๐ฅโ2)^3 (๐ฅ+1)^2 [4๐ฅ+4๐ฅ+3๐ฅโ6] = (๐ฅโ2)^3 (๐ฅ+1)^2 [7๐ฅโ2] Step 2: Putting fโ(๐ฅ)=0 (๐ฅโ2)^3 (๐ฅ+1)^2 (7๐ฅโ2)=0 (๐ฅโ2)^3 = 0 ๐ฅ โ 2 = 0 ๐ฅ=2 (๐ฅ+1)^2=0 (๐ฅ+1)=0 ๐ฅ = โ1 7๐ฅ โ 2 = 0 7๐ฅ = 2 ๐ฅ = 2/7 Hence, ๐ฅ=2 & ๐ฅ=โ1 & ๐ฅ=2/7 = 0.28 Thus ๐ฅ=โ1 is a point of inflexion ๐ฅ=2/7 is point of maxima & ๐ฅ=2 is point of minima

Miscellaneous