Last updated at April 13, 2021 by
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Ex 5.2, 9 Prove that the function f given by π (π₯) = | π₯ β 1|, π₯ β π is not differentiable at x = 1. f(x) = |π₯β1| = {β((π₯β1), π₯β1β₯0@β(π₯β1), π₯β1<0)β€ = {β((π₯β1), π₯β₯1@β(π₯β1), π₯<1)β€ Now, f(x) is a 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|(1 β β)β1|)/β = (π ππ)β¬(hβ0) (0 β|ββ|)/β = (πππ)β¬(hβ0) (0 β β)/β = (πππ)β¬(hβ0) (ββ)/β = (πππ)β¬(hβ0) (β1) = β1 (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|1 β 1|)/β = (πππ)β¬(hβ0) (|β| β 0)/β = (πππ)β¬(hβ0) (β β 0)/β = (πππ)β¬(hβ0) β/β = (πππ)β¬(hβ0) (1) = 1 Since LHD β RHD β΄ f(x) is not differentiable at x = 1 Hence proved
Checking if funciton is differentiable
Checking if funciton is differentiable
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