Example 39 - Find absolute maximum, minimum values of f(x)

Example 39 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 39 - Chapter 6 Class 12 Application of Derivatives - Part 3

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

Transcript

Example 39 Find the absolute maximum and minimum values of a function f given by ๐‘“ (๐‘ฅ) = 2๐‘ฅ3 โ€“ 15๐‘ฅ2 + 36๐‘ฅ +1 on the interval [1, 5]. f(๐‘ฅ) = 2๐‘ฅ3 โ€“ 15๐‘ฅ2 + 36๐‘ฅ + 1 Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=6๐‘ฅ^2โˆ’30๐‘ฅ+36 Putting f โ€™(๐’™)=๐ŸŽ 6๐‘ฅ2 โ€“ 30๐‘ฅ + 36 = 0 6(๐‘ฅ^2โˆ’5๐‘ฅ+6)=0 (๐‘ฅ^2โˆ’5๐‘ฅ+6)=0 (๐‘ฅ^2โˆ’2๐‘ฅโˆ’3๐‘ฅ+6)=0 ๐‘ฅ(๐‘ฅโˆ’2)โˆ’3(๐‘ฅโˆ’2)=0 (๐‘ฅโˆ’2)(๐‘ฅโˆ’3)=0 Hence. ๐’™ = 2 or ๐’™ = 3 Since we are given interval [๐Ÿ , ๐Ÿ“] Hence, calculating f(๐‘ฅ) at ๐‘ฅ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at ๐’™=๐Ÿ“ Absolute minimum value is 24 at ๐’™=๐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 10 years. He provides courses for Maths and Science at Teachoo.