Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Example 28 Find absolute maximum and minimum values of a function f given by f (đĽ) = ă12 đĽă^(4/3) â ă 6đĽă^(1/3) , đĽ â [ â 1, 1] f (đĽ) = ă12 đĽă^(4/3) â ă 6đĽă^(1/3) Finding fâ(đ) fâ(đĽ)=đ(12đĽ^(4/3) â 6đĽ^(1/3) )/đđĽ = 12 Ă 4/3 đĽ^(4/3 â1)â6 Ă 1/3 đĽ^(1/3 â1) = 4 Ă 4 đĽ^((4 â 3)/3) â2đĽ^((1 â 3)/3) = 16 đĽ^(1/3) â2đĽ^((â2)/3) = 16 đĽ^(1/3) â 2/đĽ^(2/3) = (16đĽ^(1/3) Ă đĽ^(2/3) â 2)/đĽ^(2/3) = (16đĽ^(1/3 + 2/3) â 2)/đĽ^(2/3) = (16đĽ^(3/3) â 2)/đĽ^(2/3) = (16đĽ â 2)/đĽ^(2/3) = đ(đđ â đ)" " /đ^(đ/đ) Hence, fâ(đĽ)=2(8đĽ â 1)/đĽ^(2/3) Putting fâ(đ)=đ 2(8đĽ â 1)/đĽ^(2/3) =0 2(8đĽâ1)=0 ĂđĽ^(2/3) 2(8đĽâ1)=0 8đĽâ1= 0 8đĽ=1 đ=đ/đ Note that: Since fâ(đĽ)=2(8đĽ â 1)/đĽ^(2/3) fâ(đĽ) is not defined at đ= 0 đ=đ/đ & 0 are critical points Since, we are given interval [âđ , đ] Hence calculating f(đĽ) at đĽ=âđ, 0, 1/8, đ Hence, Absolute maximum value of f(x) is 18 at đ = â1 & Absolute minimum value of f(x) is (âđ)/đ at đ = đ/đ