# Ex 6.5, 5 (iii) - 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: (iii) f(π₯) = 4π₯ β 1/2 π₯2 , π₯ β [β2, 9/2] f(π₯) = 4π₯ β 1/2 π₯2 Finding fβ(π) fβ(π₯)=π(4π₯ β 1/2 π₯^2 )/ππ₯ = 4 β 1/2 Γ2π₯ = 4 β π₯ Putting fβ(π)=π 4 β π₯=0 π₯=4 β΄ π₯=4 is only critical point Since given interval π₯ β [β2 , 9/2] Hence , calculating f(π₯) at π₯ = β 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at π = 4 & Absolute Minimum value of f(x) is β10 at π = β2

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 You are here

Ex 6.5, 5 (iv)

Ex 6.5,6

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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.