Question

Find all the points of discontinuity of f defined by f (x) = | x |− | x + 1 |.

Answer 1

Given f (x) = | x |− | x + 1 |.

The two functions, g and h, are defined as

`g(x)=|x|` and `h(x)=|x+1|`

Then ,`f=g-h`

The continuity of g and h is examined first.

`g(x)=|x|` can be written as 

`g(x)=[[-x, if   x < 0],[x, if  x ≥ 0]]`

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

`" If c < 0 , then "   g(c)=-c  and  lim_(x->c)g(x)= lim_(x->c)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x < 0.

Case II:

`"If c < 0 , then "g(c)=-c lim_(x->c)g(x)=lim_(x->c)(-x)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x > 0.

Case III:

`" if c =0 , then " g (c)=g(0)=0`

` lim_(x->0^-)g(x)= lim_(x->0^-)(- x)=0`

` lim_(x->0^+)g(x)= lim_(x->0^+)(x)=0`

` ∴ lim_(x->0^+)g(x)= lim_(x->0^+)(x)=g(0)`

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points

`h(x)=|x+1|` can be written as 

`h(x)=[[-(x+1) if   x< -1],[x+1   if x ≥ -1]]`

Clearly, h is defined for every real number.

Let c be a real number.

Case I:

`"if c < - 1, then h (c) = - (c +1) and " lim_(x->c) h (x) = lim_(x->c)[-(x+1)]=-(c + 1)`

` ∴ lim _(x-> c) h (x) = h(c) `

So, h is continuous at all points x < −1.

Case II:

`"if c > - 1, then h (c) = c +1 and " lim_(x->c) h (x) = lim _(x->c)(x + 1)= c + 1`]

` ∴lim _(x->c) h (x) = h(c)`

So, h is continuous at all points x > −1.

Case III:

if c = -1, then h (c) = h (-1) = -1+1 = 0

`lim _(x->- 1^- ) h (x) = lim_(x->-1^-)[-(x+1)]=-(-1+1)=0`

`lim _(x->- 1^+ ) h (x) = lim_(x->-1^+)( x +1)=(-1+1) =0`

`∴lim _(x->- 1^- ) h (x) = lim_(h->-1^+)= h(-1)`

So, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

So, g and h are continuous functions.

Thus, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.