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Misc 18 - If f(x) = |x|3, show that f(x) exists and find it

Misc 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.