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Question

Discuss the continuity and differentiability of f (x) = e|x| .

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Solution

Given: f(x) = ex
f(x) = ex, x0e-x, x<0

Continuity:

(LHL at x = 0)

limx0- f(x) = limh0 f(0-h) = limh0 e-(0-h) = limh0 eh = 1

(RHL at x = 0)

limx0+ f(x) = limh0 f(0+h) = limh0 e(0+h) = 1

and f(0) = e0 = 1

Thus, limx0- f(x) = limx0+ f(x) = f(0)

Hence,function is continuous at x = 0 .

Differentiability at x = 0.

(LHD at x = 0)

limx0-f(x) - f(0)x-0= limh0f(0-h) - f(0)0-h-0= limh0 e-(0-h) - 1-h= limh0 eh-1-h = -1 limh0 eh-1h =1
(RHD at x = 0)

limx0+f(x) - f(0)x-0= limh0f(0+h) - f(0)0+h-0= limh0 e(0+h) - 1h= limh0 eh-1h = 1 limh0 eh-1h =1

LHD at (x = 0)RHD at (x = 0)

Hence the function is not differentiable at x = 0.

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