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Ex 5.2, 9 - Prove that f(x) = |x - 1| is not differentiable

Ex 5.2, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.2, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.2, 9 Prove that the function f given by 𝑓 (π‘₯) = | π‘₯ – 1|, π‘₯ ∈ 𝑅 is not differentiable at x = 1. f(x) = |π‘₯βˆ’1| = {β–ˆ((π‘₯βˆ’1), π‘₯βˆ’1β‰₯0@βˆ’(π‘₯βˆ’1), π‘₯βˆ’1<0)─ = {β–ˆ((π‘₯βˆ’1), π‘₯β‰₯1@βˆ’(π‘₯βˆ’1), π‘₯<1)─ Now, f(x) is a differentiable at x = 1 if LHD = RHD (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙) βˆ’ 𝒇(𝒙 βˆ’ 𝒉))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|1 βˆ’ 1|βˆ’|(1 βˆ’ β„Ž)βˆ’1|)/β„Ž = (𝑙 π‘–π‘š)┬(hβ†’0) (0 βˆ’|βˆ’β„Ž|)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’ β„Ž)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (βˆ’β„Ž)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (βˆ’1) = βˆ’1 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|(1 + β„Ž) βˆ’ 1|βˆ’|1 βˆ’ 1|)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|β„Ž| βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (β„Ž βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) β„Ž/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (1) = 1 Since LHD β‰  RHD ∴ f(x) is not differentiable at x = 1 Hence proved

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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.