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Question
Find all the points of discontinuity of f defined by `f(x) = |x| - |x + 1|`.
Solution
`f(x) = {(-x - [-(x + 1)], if x<-1),(-(x) - (x+1), if -1 <=x<0),(x - (x+1), if x>=0):}`
`{(1, if x<-1),(-2x-1, if -1 <=x<0),(-1, if x>=0):}`
At = -1
`lim_(x->1^-) f(x) = 1`
`lim_(x->1^+) f(x) = lim_(h->0) (-2(-1+h)) = 1`
f (-1) = -2(-1) -1 = 1
Thus, `lim_(x->1^-) f (x) = lim_(x->1^+) f (x) = f (-1)`
= f is continuous at x = -1
At x= 0
`lim_(x->0^-) f(x) = lim_(x->0^-)(-2x-1) = lim_(h->0)(-2(-h)-1) = -1`
`lim_(x->0^+) f(x) = -1`
Also, f(0) = -1
Thus,`lim_(x->0^-) f(x) = lim_(x->0^+) f(x) = f(0)`
f is continuous at x = 0
Also, f being a constant is continuous when x<-1 or when x>0.
∴ f is continuous for all x ∈ R
Hence, there is no point in discontintinuty.
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