Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 You are here

Misc 19 Important

Misc 20

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Misc 23 Important

Chapter 5 Class 12 Continuity and Differentiability

Serial order wise

Last updated at May 6, 2021 by Teachoo

Misc 18 If 𝑓 (𝑥)=|𝑥|^3, show that 𝑓 ″(𝑥) exists for all real 𝑥 and find it. We know that |𝑥|={█( 𝑥 𝑥≥[email protected]−𝑥 𝑥<0)┤ Therefore, 𝑓 (𝑥)=|𝑥|^3 = {█( (𝑥)^3 , 𝑥≥[email protected](−𝑥)^3 , 𝑥<0)┤ = {█( 𝑥^3 , 𝑥≥[email protected]〖−𝑥〗^3 , 𝑥<0)┤ Case 1: When 𝒙≥𝟎 𝑓 (𝑥)=𝑥^3 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑓′(𝑥)=〖3𝑥〗^2 Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑓′′(𝑥)= (3𝑥^2 )^′ 𝒇′′(𝒙)=" " 6𝑥 Hence, 𝒇′′(𝒙) exists for all value of 𝑥 greater than 0. Case 2: When 𝒙<𝟎 𝑓 (𝑥)=〖−𝑥〗^3 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑓′(𝑥)=〖−3𝑥〗^2 Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑓′′(𝑥)= (〖−3𝑥〗^2 )^′ 𝒇^′′ (𝒙)=" "−6𝑥 Hence, 𝒇′′(𝒙) exists for all value of 𝑥 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(𝑥)= {█( 𝑥^3 , 𝑥≥[email protected]〖−𝑥〗^3 , 𝑥<0)┤ 𝒇′(𝒙)= {█( 〖3𝑥〗^2 , 𝑥≥[email protected]〖−3𝑥〗^2 , 𝑥<0)┤ We know that 𝑓′(𝑥) is differentiate at 𝑥 = 0 if L.H.D = R.H.D(𝐥𝐢𝐦)┬(𝒉 →𝟎 ) (𝒇^′ (𝟎) − 𝒇^′ (𝟎 − 𝒉))/𝒉 = lim┬(ℎ →0 ) (𝑓^′ (0) − 𝑓^′ (−ℎ))/ℎ = lim┬(ℎ →0 ) (〖3(0)〗^2 −(−〖3(−ℎ)〗^2))/ℎ = lim┬(ℎ →0 ) 〖3ℎ〗^2/ℎ = lim┬(h →0 ) (3ℎ) Putting ℎ =0 = 3(0) = 0 (𝐥𝐢𝐦)┬(𝒉 →𝟎 ) (𝒇^′ (𝟎 + 𝒉) −𝒇(𝟎))/𝒉 = lim┬(ℎ →0 ) (〖𝑓𝑎〗^′ (ℎ) − 𝑓(0))/(ℎ ) = lim┬(ℎ →0 ) (〖3(ℎ)〗^2 − 〖3(0)〗^2)/ℎ = lim┬(ℎ →0 ) 〖3ℎ〗^2/ℎ = lim┬(ℎ →0 ) 3ℎ Putting ℎ =0 = 3(0) = 0 Thus, LHD = RHD Therefore, 𝒇^′ (𝒙) is differentiable at 𝑥 = 0 So, we can say that 𝒇^′′ (𝒙) exists for x = 0 a Thus, 𝒇^′′(𝒙) exists for all real values of 𝑥 Hence proved