Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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) You are here
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
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, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (v) đ (đĽ)=đĽ3 â6đĽ2+9đĽ+15Putting fâ(đ)=đ 3(đĽ^2â4đĽ+3)=0 đĽ^2â4đĽ+3=0 đĽ^2â3đĽâđĽ+3=0 đĽ(đĽâ3)â1(đĽâ3)=0 (đĽâ1)(đĽâ3)=0 So, x = 1 & x = 3 Finding fââ(đ) fâ(đĽ)=3(đĽ^2â4đĽ+3) fââ(đĽ)=đ(3(đĽ^2 â 4đĽ+3))/đđĽ = 3(2đĽâ4+0) = 6đĽâ12 Putting đ=đ in fââ(đ) fââ(1)=6(1)â12 = 6 â 12 = â 6 < 0 Since fââ(đĽ)<0 when đĽ=1 â đĽ=1 is point of local maxima â´ f(đĽ) is maximum at đ=đ Maximum value of f(đĽ) at đĽ = 1 f(đĽ)=đĽ^3â6đĽ^2+9đĽ+15 f(1)=(1)^3â6(1)^2+9(1)+15 = 1 â 6 + 9 + 15 = 19 Putting đ=đ in fââ(x) fââ(đĽ)=6đĽâ12 fââ(3)=6(3)â12 = 18 â 12 = 6 > 0 Since fââ(đĽ)>0 when đĽ=3 â đĽ=3 is point of local minima â´ f(đĽ) is minimum at đ=đ Minimum value of f(đĽ) at đĽ = 3 f(đĽ)=đĽ^3â6đĽ^2+9đĽ+15 f(3)=(3)^3â6(3)^2+9(3)+15 = 27 â 54 + 27 + 15 = 15