# Example 40 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 40 Find absolute maximum and minimum values of a function f given by f ( ) = 12 4 3 6 1 3 , [ 1, 1] f ( ) = 12 4 3 6 1 3 Step 1: Finding f f = 12 4 3 6 1 3 = 12 4 3 4 3 1 6 1 3 1 3 1 = 4 4 4 3 3 2 1 3 3 = 16 1 3 2 2 3 = 16 1 3 2 2 3 = 16 1 3 2 3 2 2 3 = 16 1 3 + 2 3 2 2 3 = 16 3 3 2 2 3 = 16 2 2 3 = 2 8 1 2 3 Hence, f = 2 8 1 2 3 Step 2: Putting f =0 2 8 1 2 3 =0 2 8 1 =0 2 3 2 8 1 =0 8 1= 0 8 =1 = 1 8 Note that: Since f = 2 8 1 2 3 f is not defined at = 0 = 1 8 & 0 are critical points Step 3: We are given interval 1 , 1 Hence calculating f at =0, 1 8 , 1 , 1 Hence, Absolute maximum value of f(x) is 18 at = 1 & Absolute minimum value of f(x) is 9 4 at =

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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.