Check sibling questions

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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.