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


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

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.