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Example 27 - Chapter 6 Class 12 Application of Derivatives
Last updated at December 16, 2024 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Example 27 Find the absolute maximum and minimum values of a function f given by š (š„) = 2š„3 ā 15š„2 + 36š„ +1 on the interval [1, 5]. f(š„) = 2š„3 ā 15š„2 + 36š„ + 1 Finding fā(š) fā(š„)=6š„^2ā30š„+36 Putting f ā(š)=š 6š„2 ā 30š„ + 36 = 0 6(š„^2ā5š„+6)=0 (š„^2ā5š„+6)=0 (š„^2ā2š„ā3š„+6)=0 š„(š„ā2)ā3(š„ā2)=0 (š„ā2)(š„ā3)=0 Hence. š = 2 or š = 3 Since we are given interval [š , š] Hence, calculating f(š„) at š„ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at š=š Absolute minimum value is 24 at š=š
