Advertisement
Advertisement
Ex 6.5, 5 (iv) - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Aug. 19, 2021 by Teachoo
Transcript
Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f (๐ฅ) = (๐ฅ โ1)2 + 3, ๐ฅ โ [โ3,1] f (๐ฅ) = (๐ฅ โ1)2 + 3 Finding fโ(๐) fโ(x) = ๐((๐ฅ โ 1)^2+3)/๐๐ฅ = 2(๐ฅโ1) Putting fโ(๐)=๐ 2(๐ฅโ1)=0 ๐ฅโ1=0 ๐ฅ=1 Since given interval ๐ฅ โ [โ3 , 1] Hence , calculating f(๐ฅ) at ๐ฅ = โ 3 , 1 Absolute Minimum value of f(x) is 3 at ๐ = 1 & Absolute Maximum value of f(x) is 19 at ๐ = โ 3
Advertisement
Ex 6.5
Ex 6.5, 1 (i)
Important
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv) You are here
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Chapter 6 Class 12 Application of Derivatives (Term 1)
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 10 years. He provides courses for Maths and Science at Teachoo.