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