Question

Define a continuity of a function of a point.find all the points of discontinuity of defined f (x)=|x|-|x-1|

Answer 1

A function f(x) is continuous at a point x = a if
(1) lim f(x) exists which means lim f(x) = lim f(x)
x->a x->a- x->a+
(2) f(a) is defined
(3) lim f(x) = f(a)
x->a
We have f(x) = |x| - |x-1|.
Now if x < 0 then |x| = -x and |x-1| = -(x-1) = -x+1 so f(x) = -x-(-x+1)=-x+x-1=-1
So f(x) is continuous at all points x < 0
If x > 1, |x| = x and |x-1| = x-1 so f(x) = x-(x-1) = x-x+1 = 1
So f(x) is continuous at all points for x > 1.
If 0 < x < 1 then |x| = x and |x-1| = -(x-1) = -x+1. So f(x) = x -(-x+1) = x+x-1 = 2x-1.
This is a continuous function so f(x) is continuous at all poits 0 < x < 1.
We only need to examine continuity at x = 0 and x = 1 since f(x) is continuous at all other points.
Consider lim f(x). When approaching from left x < 0, so as determined above f = -1.
x->0-
So the left sided limit at x = 0 is -1.Now consider lim f(x) . When approaching from right, x
x->a+
>0 but < 1 since we are close to 0. so 0 < x < 1. We already found f(x) = 2x -1 for this range of x. So substituting x =0, value of limit = -1.
We have shown that lim f(x) = lim f(x) = -1. So lim f(x) exists and = -1. Also f() = -1
x->0- x->0+ x->0
since f(0) = |0|-|-1|= 0-1 = -1.
So f(x) is CONTINUOUS at x = 0 as it satisfies all 3 conditions for continuity at x = 0
Next we examine continuity at x = 1.
f(1) = |1|-|1-1| = 1-0 = 1
First find lim f(x)
x->1-
When approaching from left, x < 1 but > 0 since we are close to 1 so we have 0<x<1.
Hence as we found earlier, f for this range = 2x -1 . Substituting x = 1 we get 2-1 = 1.
Hence left sided limit = 1.
Next find lim f(x)
x->1+
When approaching x = 1 from right, x > 1 so the value of f = 1 and the limit = 1.
Hence we find that lim f(x) = lim f(x) =1 = lim f(x) = f(1) = 1
x->1- x = 1+ x = 1

It means f(x) is CONTINUOUS at x = 1.

So answer is f(x) is continuous at ALL points and there is no discontinuity.