Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important You are here

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26 Important

Example 27

Example 28 Important

Example 29 Important

Example 30 Important

Example 31 Important

Example 32 Important

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Deleted for CBSE Board 2024 Exams

Question 10 Deleted for CBSE Board 2024 Exams

Question 11 Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Important Deleted for CBSE Board 2024 Exams

Question 14 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Example 20 Find local maximum and local minimum values of the function f given by f (π₯)=3π₯4 + 4π₯3 β 12π₯2 + 12f (π₯)=3π₯4 + 4π₯3 β 12π₯2 + 12 Finding fβ (π) fβ (π₯)=π(3π₯4 + 4π₯3 β 12π₯2 + 12)/ππ₯ fβ (π₯)=12π₯^3+12π₯^2 β 24π₯ "+ 0" fβ (π₯)=12(π₯^3+π₯^2β2π₯) fβ (π₯)=12π₯(π₯^2+π₯β2) fβ (π₯)=12π₯ (π₯^2+2π₯βπ₯β2) fβ (π₯)=12π₯ (π₯(π₯+2)β1(π₯+2)) fβ (π₯)=πππ (πβπ)(π+π) Putting fβ (π)=π 12π₯ (π₯β1)(π₯+2)=0 π₯ (π₯β1)(π₯+2)=0 So, π=π,π₯=π,& π₯=βπ Finding fββ(π) f β(π₯)=12(π₯^3+π₯^2β2π₯) f ββ(π₯)=12π(π₯^3 + π₯^2 β 2π₯)/ππ₯ f ββ(π₯)=ππ(ππ^π+ππβπ) At π=π f ββ(0)=12(3(0)^2+2(0)β2)= 32 (0+0 β2)= β 64 < 0 Since fββ(π₯)<0 at π₯=0 β΄ π₯ = 0 is point of local maxima Thus, f(π₯) is maximum at π₯=0 At π=π fββ(1)=12(3(1)^2+2(1)β2)= 12 (3+2β2) = 36 > 0 Since fββ(π₯)>0 at π₯=1 β΄ π₯ = 1 is point of local minima Thus, f(π₯) is minimum at π₯=1 At π=βπ fββ(β2)=12(3(β2)^2+2(β2)β2)= 12 (12β4β2)= 72 > 0 Since fββ(π₯)>0 at π₯=β2 β΄ π₯ = β2 is point of local minima Thus, f(π₯) is minimum at π₯=β2 Finding local minimum and maximum value fβ (π₯)=πππ (πβπ)(π+π) Local maximum value of f (π₯) at π₯=0 f (0)=3(0)4 + 4(0)3 β 12(0)2 + 12 = 0 + 0 β 0 + 12 = 12 Local minimum value of f (π₯) at π₯=1 f (1)=3(1)4 + 4(1)3 β 12(1)2 + 12 = 3 + 4 β 12 + 12 = 7 Local Minimum value of f (π₯) at π₯=β2 f (β2)=3(β2)4 + 4(β2)3 β 12(β2)2 + 12 = 48 β 32 β 48 + 12 = β 20