Examples

Example 1
Deleted for CBSE Board 2022 Exams

Example 2 Deleted for CBSE Board 2022 Exams

Example 3 Deleted for CBSE Board 2022 Exams

Example 4 Important Deleted for CBSE Board 2022 Exams

Example 5 Deleted for CBSE Board 2022 Exams

Example 6 Deleted for CBSE Board 2022 Exams

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16

Example 17 Important

Example 18

Example 19

Example 20

Example 21 Deleted for CBSE Board 2022 Exams

Example 22 Deleted for CBSE Board 2022 Exams

Example 23 Deleted for CBSE Board 2022 Exams

Example 24 Deleted for CBSE Board 2022 Exams

Example 25 Deleted for CBSE Board 2022 Exams

Example 26

Example 27

Example 28 Important

Example 29

Example 30 Important

Example 31

Example 32 Important

Example 33 Important

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39 You are here

Example 40 Important

Example 41 Important

Example 42 Important Deleted for CBSE Board 2022 Exams

Example 43 Important Deleted for CBSE Board 2022 Exams

Example 44 Important Deleted for CBSE Board 2022 Exams

Example 45 Important Deleted for CBSE Board 2022 Exams

Example 46 Important

Example 47 Important

Example 48 Important

Example 49 Deleted for CBSE Board 2022 Exams

Example 50 Important

Example 51

Last updated at April 19, 2021 by Teachoo

Example 39 Find the absolute maximum and minimum values of a function f given by π (π₯) = 2π₯3 β 15π₯2 + 36π₯ +1 on the interval [1, 5]. f(π₯) = 2π₯3 β 15π₯2 + 36π₯ + 1 Finding fβ(π) fβ(π₯)=6π₯^2β30π₯+36 Putting f β(π)=π 6π₯2 β 30π₯ + 36 = 0 6(π₯^2β5π₯+6)=0 (π₯^2β5π₯+6)=0 (π₯^2β2π₯β3π₯+6)=0 π₯(π₯β2)β3(π₯β2)=0 (π₯β2)(π₯β3)=0 Hence. π = 2 or π = 3 Since we are given interval [π , π] Hence, calculating f(π₯) at π₯ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at π=π Absolute minimum value is 24 at π=π