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Ex 6.5, 5 (iv) - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Aug. 19, 2021 by Teachoo
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
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Ex 6.5,14 Important
Ex 6.5,15 Important
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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)
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 (π₯) = (π₯ β1)2 + 3, π₯ β [β3,1] f (π₯) = (π₯ β1)2 + 3 Finding fβ(π) fβ(x) = π((π₯ β 1)^2+3)/ππ₯ = 2(π₯β1) Putting fβ(π)=π 2(π₯β1)=0 π₯β1=0 π₯=1 Since given interval π₯ β [β3 , 1] Hence , calculating f(π₯) at π₯ = β 3 , 1 Absolute Minimum value of f(x) is 3 at π = 1 & Absolute Maximum value of f(x) is 19 at π = β 3
