Example 27 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 27 Find the absolute maximum and minimum values of a function f given by đ (đĽ) = 2đĽ3 â 15đĽ2 + 36đĽ +1 on the interval [1, 5]. f(đĽ) = 2đĽ3 â 15đĽ2 + 36đĽ + 1 Finding fâ(đ) fâ(đĽ)=6đĽ^2â30đĽ+36 Putting f â(đ)=đ 6đĽ2 â 30đĽ + 36 = 0 6(đĽ^2â5đĽ+6)=0 (đĽ^2â5đĽ+6)=0 (đĽ^2â2đĽâ3đĽ+6)=0 đĽ(đĽâ2)â3(đĽâ2)=0 (đĽâ2)(đĽâ3)=0 Hence. đ = 2 or đ = 3 Since we are given interval [đ , đ] Hence, calculating f(đĽ) at đĽ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at đ=đ Absolute minimum value is 24 at đ=đ
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo