Misc 13 Let f be a function defined on [a, b] such that f’ (𝑥) > 0, for all 𝑥 ∈ (a, b). Then prove that f is an increasing function on (a, b).We have to prove that function is always increasing i.e. f(𝒙𝟏)<𝒇(𝒙𝟐) for 𝒙𝟏 < 𝒙𝟐 where 𝒙𝟏 , 𝒙𝟐 ∈ [𝒂 , 𝒃] Proof Let 𝒙𝟏 , 𝒙𝟐 be two numbers in the interval [𝑎 , 𝑏] i.e. 𝑥1 , 𝑥2 ∈ [𝑎 , 𝑏] And, 𝒙𝟏 < 𝒙𝟐 In Interval [𝒙𝟏 ," " 𝒙𝟐] As f is defined everywhere, f is continuous & differentiable in [𝑥1 ," " 𝑥2] By Mean value of theorem, There exists c in (𝑥1 ,𝑥2) i.e. c ∈ (𝑥1 ," " 𝑥2) such that f’(c) =(𝒇(𝒙𝟐) − 𝒇(𝒙𝟏))/(𝒙𝟐 − 𝒙𝟏 ) Given that f’(𝑥)>0 for all 𝑥 ∈ (𝑎 , 𝑏) So, f’(𝒄)>𝟎 for all c ∈ (𝒙𝟏 ,𝒙𝟐) (𝒇(𝒙𝟐) − 𝒇(𝒙𝟏))/(𝒙𝟐 − 𝒙𝟏 )>𝟎 𝑓(𝑥2)−𝑓(𝑥1)>0 So, we can write that For any two points 𝑥1 , 𝑥2 in interval [𝑎 , 𝑏] Where 𝒙𝟐> 𝒙𝟏 𝒇(𝒙𝟐)> 𝒇(𝒙𝟏) Thus, f increasing in the interval [𝒂 , 𝒃] Hence proved

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.