# Ex 6.5, 5 (iv) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

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

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)

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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.