# 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

Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31 You are here

Example 32

Example 33

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41

Example 42

Example 43

Example 44

Example 45

Example 46 Important

Example 47 Important

Example 48

Example 49

Example 50

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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.