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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 (𝑥) = (𝑥 −1)2 + 3, 𝑥 ∈ [−3,1] f (𝑥) = (𝑥 −1)2 + 3 Finding f’(𝒙) f’(x) = 𝑑((𝑥 − 1)^2+3)/𝑑𝑥 = 2(𝑥−1) Putting f’(𝒙)=𝟎 2(𝑥−1)=0 𝑥−1=0 𝑥=1 Since given interval 𝑥 ∈ [−3 , 1] Hence , calculating f(𝑥) at 𝑥 = – 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.