Check sibling questions

Ex 5.2, 9 - Chapter 5 Class 12 Continuity and Differentiability (Important Question)

Last updated at May 29, 2023 by

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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.