Example 27 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

  Example 27 - Find absolute maximum, minimum values of f(x) - Examples

part 2 - Example 27 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Example 27 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Example 27 Find the absolute maximum and minimum values of a function f given by 𝑓 (𝑥) = 2𝑥3 – 15𝑥2 + 36𝑥 +1 on the interval [1, 5]. f(𝑥) = 2𝑥3 – 15𝑥2 + 36𝑥 + 1 Finding f’(𝒙) f’(𝑥)=6𝑥^2−30𝑥+36 Putting f ’(𝒙)=𝟎 6𝑥2 – 30𝑥 + 36 = 0 6(𝑥^2−5𝑥+6)=0 (𝑥^2−5𝑥+6)=0 (𝑥^2−2𝑥−3𝑥+6)=0 𝑥(𝑥−2)−3(𝑥−2)=0 (𝑥−2)(𝑥−3)=0 Hence. 𝒙 = 2 or 𝒙 = 3 Since we are given interval [𝟏 , 𝟓] Hence, calculating f(𝑥) at 𝑥 = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at 𝒙=𝟓 Absolute minimum value is 24 at 𝒙=𝟏

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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