Ex 6.5, 1 - Find maximum and minimum values of (i) f(x) =

Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 6
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 7
Ex 6.5,1 - Chapter 6 Class 12 Application of Derivatives - Part 8

 

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐‘ฅ) = (2๐‘ฅ โ€“ 1)^2 + 3f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Hence, Minimum value of (2๐‘ฅโˆ’1)^2 = 0 Minimum value of (2๐‘ฅโˆ’1^2 )+3 = 0 + 3 = 3 Square of number cant be negative It can be 0 or greater than 0 Also, there is no maximum value of ๐‘ฅ โˆด There is no maximum value of f(x) Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐‘ฅ) = (2๐‘ฅ โ€“ 1)^2+3Finding fโ€™(x) f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 fโ€™(๐‘ฅ)= 2(2๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ 2(2๐‘ฅโˆ’1)=0 2๐‘ฅ โ€“ 1 = 0 2๐‘ฅ = 1 ๐‘ฅ = 1/2 Thus, x = 1/2 is the minima Finding minimum value f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Putting ๐‘ฅ = 1/2 f(1/2)=(2 ร— 1/2โˆ’1)^2+3= (1โˆ’1)^2+3= 3 โˆด Minimum value = 3 There is no maximum value Ex 6.5, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) ๐‘“ (๐‘ฅ)= (2๐‘ฅ โ€“ 1)^2 + 3Double Derivative Test f(๐‘ฅ)=(2๐‘ฅโˆ’1)^2+3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=2(2๐‘ฅโˆ’1)^(2โˆ’1) = 2(2๐‘ฅโˆ’1) Putting fโ€™(๐’™)=๐ŸŽ 2(2๐‘ฅโˆ’1)=0 (2๐‘ฅโˆ’1)=0 2๐‘ฅ = 0 + 1 ๐‘ฅ = 1/2 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=2(2๐‘ฅโˆ’1) fโ€™(๐‘ฅ) = 4๐‘ฅ โ€“ 2 fโ€™โ€™(๐‘ฅ)= 4 So, fโ€™โ€™ (1/2) = 4 Since fโ€™โ€™ (1/2) > 0 , ๐‘ฅ = 1/2 is point of local minima Putting ๐‘ฅ = 1/2 , we can calculate minimum value f(๐‘ฅ) = (2๐‘ฅโˆ’1)^2+3 f(1/2)= (2(1/2)โˆ’1)^2+3= (1โˆ’1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.