# Example 31 - 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 31 (Method 1) Find local minimum value of the function f given by f ( ) = 3 + | |, R. f ( ) = 3 + | | R Since Value of 0 So, Minimum value of =0 Minimum value of f ( ) = 3 + Minimum value of = 3 + 0 = 3 Hence minimum value of f ( ) = 3 We cannot find any maximum value. Example 31 (Method 2) Find local minimum value of the function f given by f ( ) = 3 + | |, R. f ( ) = 3 + | | We know that = & , <0 & , 0 Hence, f ( ) = 3 & , <0 3+ , 0 = 0 is only critical point We check sign of f ( ) at = 0 Since sign of f (x) changes from negative to positive, x = 0 is point of Maxima & f ( ) is Minimum at = 0 Minimum value of f ( ) f (0) = 3 + |0|= 3 There s no Maximum value of f ( ) Example 31 (Method 3) Find local minimum value of the function f given by f ( ) = 3 + | |, R. f ( ) = 3 + | | From graph, f ( ) is Minimum at = 0 So, Minimum value of f ( ) = f (0) = 3

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Example 31 You are here

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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.