Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Example 28 Find absolute maximum and minimum values of a function f given by f (π₯) = γ12 π₯γ^(4/3) β γ 6π₯γ^(1/3) , π₯ β [ β 1, 1] f (π₯) = γ12 π₯γ^(4/3) β γ 6π₯γ^(1/3) Finding fβ(π) fβ(π₯)=π(12π₯^(4/3) β 6π₯^(1/3) )/ππ₯ = 12 Γ 4/3 π₯^(4/3 β1)β6 Γ 1/3 π₯^(1/3 β1) = 4 Γ 4 π₯^((4 β 3)/3) β2π₯^((1 β 3)/3) = 16 π₯^(1/3) β2π₯^((β2)/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(ππ 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₯β1)=0 Γπ₯^(2/3) 2(8π₯β1)=0 8π₯β1= 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 [βπ , π] Hence calculating f(π₯) at π₯=βπ, 0, 1/8, π f(0)=12(0)^(4/3)β6(0)^(1/3) = 0 β 0 = 0 f(1)=12(1)^(4/3)β6(1)^(1/3) = 12 β 6 = 6 Hence, Absolute maximum value of f(x) is 18 at π = β1 & Absolute minimum value of f(x) is (βπ)/π at π = π/π

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.