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Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 9

Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 10
Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives - Part 11

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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‘₯ βˆ’1)2 + 3, π‘₯ ∈ [βˆ’3,1] f (π‘₯) = (π‘₯ βˆ’1)2 + 3 Finding f’(𝒙) f’(x) = 𝑑((π‘₯ βˆ’ 1)^2+3)/𝑑π‘₯ = 2(π‘₯βˆ’1) Putting f’(𝒙)=𝟎 2(π‘₯βˆ’1)=0 π‘₯βˆ’1=0 π‘₯=1 Since given interval π‘₯ ∈ [βˆ’3 , 1] Hence , calculating f(π‘₯) at π‘₯ = – 3 , 1 Absolute Minimum value of f(x) is 3 at 𝒙 = 1 & Absolute Maximum value of f(x) is 19 at 𝒙 = – 3

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.