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Question
I am troubled in this question.
If f(x) = │x-1│-│x│ then discuss continuity and differentiability of f(x) at x=0 and x=1

Thanks in Advance

Answer
Continuity
---------------------
The value is continuous everywhere.
The points to check to make sure this is true are when the values inside | | change sign.

The left and right limits as x->1 of |x-1| are both 0, and so is the value at x=1.

The left and right limits as x->0 of |x| are both 0 and so is the value at x=0.

Thus, the function is continous.

The function is then
f(x) = 1 for x < 0,
f(x) = 1 - 2x 0<x<1, and
f(x) = 1 for x > 1

Differentiability
---------------------
f'(x) = 0 for x < 0,
f'(x) = -2 for < x < 1, and
f'(x) = 0 for 1 < x.

Thus at the points 0 and 1, the derivative on one outside is 0 and on the inside, it's -2.
Therefore at the points x=0 and x=1, the derivative does not exist.

Calculus

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