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Example 40 - Find absolute max, min values of f(x) = 12x4/3

Example 40 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 40 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 40 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 40 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 40 - Chapter 6 Class 12 Application of Derivatives - Part 6

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Example 40 Find absolute maximum and minimum values of a function f given by f (π‘₯) = γ€–12 π‘₯γ€—^(4/3) – γ€– 6π‘₯γ€—^(1/3) , π‘₯ ∈ [ – 1, 1] f (π‘₯) = γ€–12 π‘₯γ€—^(4/3) – γ€– 6π‘₯γ€—^(1/3) Finding f’(𝒙) f’(π‘₯)=𝑑(12π‘₯^(4/3) βˆ’ 6π‘₯^(1/3) )/𝑑π‘₯ = 12 Γ— 4/3 π‘₯^(4/3 βˆ’1)βˆ’6 Γ— 1/3 π‘₯^(1/3 βˆ’1) = 4 Γ— 4 π‘₯^((4 βˆ’ 3)/3) βˆ’2π‘₯^((1 βˆ’ 3)/3) = 16 π‘₯^(1/3) βˆ’2π‘₯^((βˆ’2)/3) = 16 π‘₯^(1/3) βˆ’ 2/π‘₯^(2/3) = (16π‘₯^(1/3) Γ— π‘₯^(2/3) βˆ’ 2)/π‘₯^(2/3) = (16π‘₯^(1/3 + 2/3) βˆ’ 2)/π‘₯^(2/3) = (16π‘₯^(3/3) βˆ’ 2)/π‘₯^(2/3) = (16π‘₯ βˆ’ 2)/π‘₯^(2/3) = 𝟐(πŸ–π’™ βˆ’ 𝟏)" " /𝒙^(𝟐/πŸ‘) Hence, f’(π‘₯)=2(8π‘₯ βˆ’ 1)/π‘₯^(2/3) Putting f’(𝒙)=𝟎 2(8π‘₯ βˆ’ 1)/π‘₯^(2/3) =0 2(8π‘₯βˆ’1)=0 Γ—π‘₯^(2/3) 2(8π‘₯βˆ’1)=0 8π‘₯βˆ’1= 0 8π‘₯=1 𝒙=𝟏/πŸ– Note that: Since f’(π‘₯)=2(8π‘₯ βˆ’ 1)/π‘₯^(2/3) f’(π‘₯) is not defined at 𝒙= 0 𝒙=𝟏/πŸ– & 0 are critical points Since, we are given interval [βˆ’πŸ , 𝟏] Hence calculating f(π‘₯) at π‘₯=βˆ’πŸ, 0, 1/8, 𝟏 f(0)=12(0)^(4/3)βˆ’6(0)^(1/3) = 0 – 0 = 0 f(1)=12(1)^(4/3)βˆ’6(1)^(1/3) = 12 – 6 = 6 Hence, Absolute maximum value of f(x) is 18 at 𝒙 = –1 & Absolute minimum value of f(x) is (βˆ’πŸ—)/πŸ’ at 𝒙 = 𝟏/πŸ–

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.