Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 6

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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 7

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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 8

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(π‘₯) = 4π‘₯ – 1/2 π‘₯2 , π‘₯ ∈ [βˆ’2, 9/2] f(π‘₯) = 4π‘₯ – 1/2 π‘₯2 Finding f’(𝒙) f’(π‘₯)=𝑑(4π‘₯ βˆ’ 1/2 π‘₯^2 )/𝑑π‘₯ = 4 – 1/2 Γ—2π‘₯ = 4 – π‘₯ Putting f’(𝒙)=𝟎 4 – π‘₯=0 π‘₯=4 ∴ π‘₯=4 is only critical point Since given interval π‘₯ ∈ [βˆ’2 , 9/2] Hence , calculating f(π‘₯) at π‘₯ = – 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at 𝒙 = 4 & Absolute Minimum value of f(x) is –10 at 𝒙 = –2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.