Ex 6.5, 5 (i) - 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: (i) f (๐ฅ) = ๐ฅ3, ๐ฅโ[โ 2, 2]Finding fโ(๐) fโ(๐ฅ)=๐(๐ฅ^3 )/๐๐ฅ fโ(๐ฅ)=3๐ฅ^2 Putting fโ(๐)=๐ 3๐ฅ^2=0 ๐ฅ^2=0 ๐ฅ=0 So, ๐ฅ=0 is critical point Since given interval is ๐ฅ โ [โ2 , 2] Hence calculating f(๐ฅ) at ๐ฅ=โ2 , 0 , 2 Absolute Maximum value of f(x) is 8 at ๐ = 2 & Absolute Minimum value of f(x) is โ8 at ๐ = โ2
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) You are here
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
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Ex 6.5,12 Important
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Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
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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)
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