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Ex 6.3, 5 (iv) - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2023 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
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Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv) You are here
Ex 6.3,6
Ex 6.3,7 Important
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Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Transcript
Ex 6.3, 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
