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Ex 6.2, 18 - Prove that f(x) = x3 - 3x2 + 3x - 100 is increasing

Ex 6.2,18 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Ex 6.2, 18 Prove that the function given by f (𝑥) = 𝑥3 – 3𝑥2 + 3𝑥 – 100 is increasing in R. We need to show f(𝑥) is strictly increasing on R i.e. we need to show f’(𝒙) > 0 Finding f’(𝒙) f’(𝑥)= 3x2 – 6x + 3 – 0 = 3(𝑥2−2𝑥+1) = 3((𝑥)2+(1)2−2(𝑥)(1)) = 3(𝑥−1)2 Since Square of any number is always positive (𝑥−1)2 > 0 3(𝑥−1)2>0 f’(𝒙) > 0 Hence, f’(𝑥) > 0 for all values of 𝑥 ∴ f(𝑥) is strictly increasing on R

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.