1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Ex 6.3
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Ex 6.3, 5 (iv) - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

 

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
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Ex 6.3

    Examples →
Examples →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo