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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Misc 18 If 𝑓 (π‘₯)=|π‘₯|^3, show that 𝑓 β€³(π‘₯) exists for all real π‘₯ and find it. We know that |π‘₯|={β–ˆ( π‘₯ π‘₯β‰₯0@βˆ’π‘₯ π‘₯<0)─ Therefore, 𝑓 (π‘₯)=|π‘₯|^3 = {β–ˆ( (π‘₯)^3 , π‘₯β‰₯0@(βˆ’π‘₯)^3 , π‘₯<0)─ = {β–ˆ( π‘₯^3 , π‘₯β‰₯0@γ€–βˆ’π‘₯γ€—^3 , π‘₯<0)─ Case 1: When 𝒙β‰₯𝟎 𝑓 (π‘₯)=π‘₯^3 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑓′(π‘₯)=γ€–3π‘₯γ€—^2 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑓′′(π‘₯)= (3π‘₯^2 )^β€² 𝒇′′(𝒙)=" " 6π‘₯ Hence, 𝒇′′(𝒙) exists for all value of π‘₯ greater than 0. Case 2: When 𝒙<𝟎 𝑓 (π‘₯)=γ€–βˆ’π‘₯γ€—^3 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑓′(π‘₯)=γ€–βˆ’3π‘₯γ€—^2 Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑓′′(π‘₯)= (γ€–βˆ’3π‘₯γ€—^2 )^β€² 𝒇^β€²β€² (𝒙)=" "βˆ’6π‘₯ Hence, 𝒇′′(𝒙) exists for all value of π‘₯ less than 0. Case 3: At x = 0 To check if 𝒇′′(𝒙) exists for x = 0, We need to check differentiability of 𝒇′(𝒙) at 𝒙 = 𝟎 Here, 𝑓(π‘₯)= {β–ˆ( π‘₯^3 , π‘₯β‰₯0@γ€–βˆ’π‘₯γ€—^3 , π‘₯<0)─ 𝒇′(𝒙)= {β–ˆ( γ€–3π‘₯γ€—^2 , π‘₯β‰₯0@γ€–βˆ’3π‘₯γ€—^2 , π‘₯<0)─ We know that 𝑓′(π‘₯) is differentiate at π‘₯ = 0 if L.H.D = R.H.D(π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇^β€² (𝟎) βˆ’ 𝒇^β€² (𝟎 βˆ’ 𝒉))/𝒉 = lim┬(β„Ž β†’0 ) (𝑓^β€² (0) βˆ’ 𝑓^β€² (βˆ’β„Ž))/β„Ž = lim┬(β„Ž β†’0 ) (γ€–3(0)γ€—^2 βˆ’(βˆ’γ€–3(βˆ’β„Ž)γ€—^2))/β„Ž = lim┬(β„Ž β†’0 ) γ€–3β„Žγ€—^2/β„Ž = lim┬(h β†’0 ) (3β„Ž) Putting β„Ž =0 = 3(0) = 0 (π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇^β€² (𝟎 + 𝒉) βˆ’π’‡(𝟎))/𝒉 = lim┬(β„Ž β†’0 ) (γ€–π‘“π‘Žγ€—^β€² (β„Ž) βˆ’ 𝑓(0))/(β„Ž ) = lim┬(β„Ž β†’0 ) (γ€–3(β„Ž)γ€—^2 βˆ’ γ€–3(0)γ€—^2)/β„Ž = lim┬(β„Ž β†’0 ) γ€–3β„Žγ€—^2/β„Ž = lim┬(β„Ž β†’0 ) 3β„Ž Putting β„Ž =0 = 3(0) = 0 Thus, LHD = RHD Therefore, 𝒇^β€² (𝒙) is differentiable at π‘₯ = 0 So, we can say that 𝒇^β€²β€² (𝒙) exists for x = 0 a Thus, 𝒇^β€²β€²(𝒙) exists for all real values of π‘₯ Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.