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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


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‘₯ 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.