Last updated at April 13, 2021 by

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Ex 5.2, 10 (Introduction) Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at π₯=1 and π₯= 2. Ex 5.2, 10 Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at π₯=1 and π₯= 2. f (x) = [x] Letβs check for both x = 1 and x = 2 At x = 1 f (x) is differentiable at x = 1 if LHD = RHD (πππ)β¬(π‘βπ) (π(π) β π(π β π))/π = (πππ)β¬(hβ0) (π(1) β π(1 β β))/β = (πππ)β¬(hβ0) ([1] β [(1 β β)])/β = (πππ)β¬(hβ0) (1 β 0)/β = (πππ)β¬(hβ0) 1/β = 1/0 = Not defined (πππ)β¬(π‘βπ) (π(π + π) β π(π))/π = (πππ)β¬(hβ0) (π(1 + β) β π(1))/β = (πππ)β¬(hβ0) ([(1 + β)] β [1])/β = (πππ)β¬(hβ0) (1 β 1)/β = (πππ)β¬(hβ0) 0/β = (πππ)β¬(hβ0) 0 = 0 For greatest integer function [1] = 1 [1 β h] = 0 [1 + h] = 1 Since LHD β RHD β΄ f(x) is not differentiable at x = 1 Hence proved At x = 2 f (x) is differentiable at x = 2 if LHD = RHD L H D (πππ)β¬(π‘βπ) (π(π) β π(π β π))/π = (πππ)β¬(hβ0) (π(2) β π(2 β β))/β = (πππ)β¬(hβ0) ([2] β [(2 β β)])/β = (πππ)β¬(hβ0) (2 β 1)/β = (πππ)β¬(hβ0) 1/β = 1/0 = Not defined R H D (πππ)β¬(π‘βπ) (π(π + π) β π(π))/π = (πππ)β¬(hβ0) (π(2 + β) β π(2))/β = (πππ)β¬(hβ0) ([(2 + β)] β [2])/β = (πππ)β¬(hβ0) (2 β 2)/β = (πππ)β¬(hβ0) 0/β = (πππ)β¬(hβ0) 0 = 0 For greatest integer function [2] = 2 [2 β h] = 1 [2 + h] = 2 Since LHD β RHD β΄ f(x) is not differentiable at x = 2 Hence proved

About the Author

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.