Example 27 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

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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