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Chapter 6 Class 12 Application of Derivatives

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Last updated at April 19, 2021 by Teachoo

Example 40 Find absolute maximum and minimum values of a function f given by f (π₯) = γ12 π₯γ^(4/3) β γ 6π₯γ^(1/3) , π₯ β [ β 1, 1] f (π₯) = γ12 π₯γ^(4/3) β γ 6π₯γ^(1/3) Finding fβ(π) fβ(π₯)=π(12π₯^(4/3) β 6π₯^(1/3) )/ππ₯ = 12 Γ 4/3 π₯^(4/3 β1)β6 Γ 1/3 π₯^(1/3 β1) = 4 Γ 4 π₯^((4 β 3)/3) β2π₯^((1 β 3)/3) = 16 π₯^(1/3) β2π₯^((β2)/3) = 16 π₯^(1/3) β 2/π₯^(2/3) = (16π₯^(1/3) Γ π₯^(2/3) β 2)/π₯^(2/3) = (16π₯^(1/3 + 2/3) β 2)/π₯^(2/3) = (16π₯^(3/3) β 2)/π₯^(2/3) = (16π₯ β 2)/π₯^(2/3) = π(ππ β π)" " /π^(π/π) Hence, fβ(π₯)=2(8π₯ β 1)/π₯^(2/3) Putting fβ(π)=π 2(8π₯ β 1)/π₯^(2/3) =0 2(8π₯β1)=0 Γπ₯^(2/3) 2(8π₯β1)=0 8π₯β1= 0 8π₯=1 π=π/π Note that: Since fβ(π₯)=2(8π₯ β 1)/π₯^(2/3) fβ(π₯) is not defined at π= 0 π=π/π & 0 are critical points Since, we are given interval [βπ , π] Hence calculating f(π₯) at π₯=βπ, 0, 1/8, π f(0)=12(0)^(4/3)β6(0)^(1/3) = 0 β 0 = 0 f(1)=12(1)^(4/3)β6(1)^(1/3) = 12 β 6 = 6 Hence, Absolute maximum value of f(x) is 18 at π = β1 & Absolute minimum value of f(x) is (βπ)/π at π = π/π