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Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important You are here
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at June 12, 2023 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