# Example 39

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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Example 39 You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .