Misc 20 - Chapter 5 Class 12 Continuity and Differentiability

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Misc 20 Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.Consider the function 𝑓(π‘₯)=|π‘₯|+|π‘₯βˆ’1| 𝑓 is continuous everywhere , but it is not differentiable at π‘₯ = 0 & π‘₯ = 1 𝑓(π‘₯)={β–ˆ( βˆ’π‘₯βˆ’(π‘₯βˆ’1) π‘₯≀0@π‘₯βˆ’(π‘₯βˆ’1) 0<π‘₯<1@π‘₯+(π‘₯βˆ’1) π‘₯β‰₯1)─ = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking both continuity and differentiability at For x < 0 For x > 1 For 0 < x < 1 For x = 0 For x = 1 Case 1: For π‘₯<0 𝑓(π‘₯)=βˆ’2π‘₯+1 𝑓(π‘₯) is polynomial ∴ 𝑓(π‘₯) is continuous & differentiable Case 2: For π‘₯>1 𝑓(π‘₯)=2π‘₯βˆ’1 𝑓(π‘₯) is polynomial ∴ 𝑓(π‘₯) is continuous & differentiable Case 3: For 0<π‘₯<1 𝑓(π‘₯)=1 𝑓(π‘₯) is a constant function ∴ 𝑓(π‘₯) is continuous & differentiable Case 4: At π‘₯=0 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking continuity A function is continuous at π‘₯=0 if LHL = RHL = 𝑓(0) i.e. (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^βˆ’ ) 𝒇(𝒙) = (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^+ ) 𝒇(𝒙) = 𝑓(0) (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^βˆ’ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(0βˆ’β„Ž) =lim┬(β„Ž β†’0) 𝑓 (βˆ’β„Ž) = lim┬(β„Ž β†’0) βˆ’2(βˆ’β„Ž)+1 = 1 (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^+ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(0+β„Ž) = lim┬(β„Ž β†’0) 𝑓(β„Ž) = lim┬(β„Ž β†’0) 1 = 1 And, 𝑓(0)= βˆ’2(0)+1= 1 Hence, LHL = RHL = f (0) ∴ 𝑓 is continuous Checking Differentiability at x = 0 𝑓 is differentiable at π‘₯ = 0 if L.H.D = R.H.D i.e., lim┬(β„Ž β†’0 ) (𝑓(0) βˆ’ 𝑓(0 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’0 ) (𝑓(0 + β„Ž) βˆ’ 𝑓(0))/β„Ž (π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇(𝟎) βˆ’ 𝒇(𝟎 βˆ’ 𝒉))/𝒉 =lim┬(β„Ž β†’0 ) (𝑓(0) βˆ’ 𝑓(βˆ’β„Ž))/(β„Ž ) =lim┬(β„Ž β†’0) ((βˆ’2(0) + 1) βˆ’ (2(βˆ’β„Ž)+1))/β„Ž =lim┬(β„Ž β†’0 ) (1 + 2β„Ž βˆ’1)/β„Ž =lim┬(β„Ž β†’0 ) 2β„Ž/β„Ž =lim┬(β„Ž β†’0) 2 = 2 (π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇(𝟎 + 𝒉) βˆ’ 𝒇(𝟎))/𝒉 =lim┬(β„Žβ†’0) (𝑓 (β„Ž) βˆ’ 𝑓(0))/β„Ž =lim┬(β„Žβ†’0 ) (1 βˆ’ (βˆ’2(0) + 1))/β„Ž =lim┬(β„Žβ†’0 ) (1 βˆ’ 1)/β„Ž =lim┬(β„Žβ†’0) 0/β„Ž = 0 Since L.H.D β‰  R.H.D ∴ 𝑓(π‘₯) is not differentiable at π‘₯=0 Case 5: At π‘₯=1 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking continuity A function is continuous at π‘₯=1 if LHL = RHL = 𝑓(1) i.e. (π₯𝐒𝐦)┬(𝒙 β†’πŸ^βˆ’ ) 𝒇(𝒙) = (π₯𝐒𝐦)┬(𝒙 β†’πŸ^+ ) 𝒇(𝒙) = 𝑓(1) (π₯𝐒𝐦)┬(𝒙 β†’πŸ^βˆ’ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(1βˆ’β„Ž) =lim┬(β„Ž β†’0) 1 = 1 (π₯𝐒𝐦)┬(𝒙 β†’πŸ^+ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(1+β„Ž) = lim┬(β„Ž β†’0) 2(1+β„Ž)βˆ’1 = 2(1 + 0) βˆ’ 1 = 1 And, 𝑓(0)= 2(1)βˆ’1= 1 Hence, LHL = RHL = f (1) ∴ 𝑓 is continuous Checking Differentiability at x = 1 𝑓 is differentiable at π‘₯ =1 if L.H.D = R.H.D i.e., lim┬(β„Ž β†’0 ) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’0 ) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž lim┬(β„Ž β†’0 ) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž) )/β„Ž =lim┬(β„Ž β†’0 ) ((2(1) βˆ’ 1) βˆ’ 1)/(β„Ž ) =lim┬(β„Ž β†’0 ) (1 βˆ’ 1)/(β„Ž ) =lim┬(β„Ž β†’0 ) 0/β„Ž = 0 lim┬(β„Ž β†’0 ) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž =lim┬(β„Ž β†’0 ) ((2(1 + β„Ž) βˆ’ 1) βˆ’ (2(1) βˆ’1))/β„Ž = lim┬(β„Ž β†’0 ) ((2 + 2β„Ž βˆ’1) βˆ’ (2 βˆ’1))/β„Ž = lim┬(β„Ž β†’0 ) ((1 + 2β„Ž) βˆ’ 1)/β„Ž = lim┬(β„Ž β†’0 ) 2β„Ž/β„Ž =lim┬(β„Ž β†’0 ) 2 = 2Since L.H.D β‰  R.H.D ∴ 𝑓 is not differentiable at π‘₯=1 Thus , 𝑓 is not differentiable at π‘₯=0 & π‘₯=1 , but continuous at all points Note :- Here we can take function |𝒙|=|π’™βˆ’π’‚|+|π’™βˆ’π’ƒ| where a & b can have any constant value . 𝑓 will be continuous at all points , but 𝑓 is not differentiable at 𝒙=𝒂 & 𝒙=𝒃

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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 and Computer Science at Teachoo