Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Example 39 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
Transcript
Example 39 Find the absolute maximum and minimum values of a function f given by đ (đ„) = 2đ„3 â 15đ„2 + 36đ„ +1 on the interval [1, 5]. fï·đ„ï·Ż = 2đ„3 â 15đ„2 + 36đ„ + 1 fâï·đ„ï·Ż=6ï·đ„ï·ź2ï·Żâ30đ„+36 Putting f âï·đ„ï·Ż=0 6đ„2 â 30đ„ + 36 = 0 6ï·ï·đ„ï·ź2ï·Żâ5đ„+6ï·Ż=0 ï·ï·đ„ï·ź2ï·Żâ5đ„+6ï·Ż=0 ï·ï·đ„ï·ź2ï·Żâ2đ„â3đ„+6ï·Ż=0 đ„ï·đ„â2ï·Żâ3ï·đ„â2ï·Ż=0 ï·đ„â2ï·Żï·đ„â3ï·Ż=0 Hence đ„ = 2 or đ„ = 3 We are given interval ï·1 , 5ï·Ż Hence calculating fï·đ„ï·Ż at đ„ = 2, 3, 1, 5 Hence, Absolute maximum value is 56 at đ=đ Absolute minimum value is 24 at đ=đ
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Chapter 6 Class 12 Application of Derivatives
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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 9 years. He provides courses for Maths and Science at Teachoo.