Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at Jan. 7, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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 (π₯) = π₯3, π₯β[β 2, 2] Finding fβ(π) fβ(π₯)=π(π₯^3 )/ππ₯ fβ(π₯)=3π₯^2 Putting fβ(π)=π 3π₯^2=0 π₯^2=0 π₯=0 So, π₯=0 is critical point Since given interval is π₯ β [β2 , 2] Hence calculating f(π₯) at π₯=β2 , 0 , 2 Absolute Maximum value of f(x) is 8 at π = 2 & Absolute Minimum value of f(x) is β8 at π = β2 Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (π₯) = sinβ‘π₯ + cosβ‘π₯ , π₯ β [0, π ] Finding fβ(π) fβ(π₯)=π(π πππ₯ + πππ π₯)/ππ₯ fβ(π₯)=cosβ‘γπ₯ βsinβ‘π₯ γ Putting fβ(π) cosβ‘γπ₯ βsinβ‘π₯ γ= 0 cosβ‘γπ₯=sinβ‘π₯ γ 1 = sinβ‘π₯/(cosβ‘ π₯) 1 = tan π₯ tan π₯ = 1 We know that know tan ΞΈ = 1 at ΞΈ = π/4 β΄ π₯ = π/4 Since given interval π₯ β [0 , π] Hence calculating f(π₯) at π₯=0 , π/4 ,π Absolute Maximum value of f(π₯) is βπ at π = π /π & Absolute Minimum value of f(π₯) is β1 at π = Ο 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, 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

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.