Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Absolute minima/maxima
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 7 Find both the maximum value and the minimum value of 3đĽ4 â 8đĽ3 + 12đĽ2 â 48đĽ + 25 on the interval [0, 3].Let f(x) = 3đĽ4 â 8đĽ3 + 12đĽ2 â 48đĽ + 25, where đĽ â [0, 3] Finding fâ(đ) fâ(đĽ)=đ(3đĽ^4 â 8đĽ^3 + 12đĽ^2 â 48đĽ + 25)/đđĽ fâ(đĽ)=3 Ă4đĽ^3â8 Ă3đĽ^2+12 Ă2đĽâ48+0 fâ(đĽ)=12đĽ^3â24đĽ^2+24đĽâ48 fâ(đĽ)=12(đĽ^3â2đĽ^2+2đĽâ4) Putting fâ(đ)=đ 12(đĽ^3â2đĽ^2+2đĽâ4)=0 đĽ^3â2đĽ^2+2đĽâ4=0 đĽ^2 (đĽâ2)+2(đĽâ2)=0 (đĽ^2+2)(đĽâ2)=0 Since đĽ^2=â2 is not possible Thus đĽ=2 is only critical point Since are given interval đĽ â [0 , 3] Hence , calculating f(đĽ) at đĽ = 0 , 2 & 3 Hence, Minimum value of f(đĽ) is â39 at đ = 2 Maximum value of f(đĽ) is 25 at đ = 0