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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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