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

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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| f(x) is a differentiable at x = 1 if LHD = RHD L H D (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(π‘₯) βˆ’ 𝑓(π‘₯ βˆ’ β„Ž))/β„Ž = (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 = (π‘™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 R H D (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯))/β„Ž = (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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.