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Ex 6.2, 1 Class 12 - Show that f(x) = 3x + 17 is strictly increasing

Ex 6.2, 1 - Chapter 6 Class 12 Application of Derivatives - Part 2


Transcript

Ex 6.2, 1 (Method 1) Show that the function given by f (𝑥) = 3𝑥 + 17 is strictly increasing on R. f(𝑥) = 3𝑥 + 17 Finding f’(𝒙) f’(𝑥) = 3 Since f’(𝒙) > 0 Hence, f is strictly increasing on R Ex 6.2, 1 (Method 2) Show that the function given by f (x) = 3x + 17 is strictly increasing on R. Let 𝑥1 and 𝑥2 be real numbers Such that 𝒙𝟏 < 𝒙2 Multiplying both sides by 3 3𝑥1 < 3 𝑥2 Adding both sides by 17 3𝑥1 + 17 < 3𝑥2 + 17 f (𝒙𝟏) < f ( 𝒙2) Hence, when x1 < x2 , f(x1) < f(x2) Thus, 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.