Ex 6.5, 5 - Find absolute maximum and min value of - Class 12

Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 2

 

 

  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: (i) f (๐‘ฅ) = ๐‘ฅ3, ๐‘ฅโˆˆ[โ€“ 2, 2]Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=3๐‘ฅ^2 Putting fโ€™(๐’™)=๐ŸŽ 3๐‘ฅ^2=0 ๐‘ฅ^2=0 ๐‘ฅ=0 So, ๐‘ฅ=0 is critical point Since given interval is ๐‘ฅ โˆˆ [โˆ’2 , 2] Hence calculating f(๐‘ฅ) at ๐‘ฅ=โˆ’2 , 0 , 2 Absolute Maximum value of f(x) is 8 at ๐’™ = 2 & Absolute Minimum value of f(x) is โ€“8 at ๐’™ = โ€“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.