Example 39 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
Transcript
Example 39 Find the absolute maximum and minimum values of a function f given by ๐ (๐ฅ) = 2๐ฅ3 โ 15๐ฅ2 + 36๐ฅ +1 on the interval [1, 5]. f(๐ฅ) = 2๐ฅ3 โ 15๐ฅ2 + 36๐ฅ + 1 Finding fโ(๐) fโ(๐ฅ)=6๐ฅ^2โ30๐ฅ+36 Putting f โ(๐)=๐ 6๐ฅ2 โ 30๐ฅ + 36 = 0 6(๐ฅ^2โ5๐ฅ+6)=0 (๐ฅ^2โ5๐ฅ+6)=0 (๐ฅ^2โ2๐ฅโ3๐ฅ+6)=0 ๐ฅ(๐ฅโ2)โ3(๐ฅโ2)=0 (๐ฅโ2)(๐ฅโ3)=0 Hence. ๐ = 2 or ๐ = 3 Since we are given interval [๐ , ๐] Hence, calculating f(๐ฅ) at ๐ฅ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at ๐=๐ Absolute minimum value is 24 at ๐=๐
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About the Author
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.