Question

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

Answer 1

We have

f(x) = `{(-x - [-(x + 1)]", if"  x<-1),(-(x) - (x+1)", if" -1 <=x<0),(x - (x+1)", if"  x>=0):}`

f(x) = `{(1", if"  x<-1),(-2x-1", if" -1 <=x<0),(-1", if"  x>=0):}`

At x = −1

`lim_(x->1^-)` f(x) = 1

`lim_(x->1^+)` f(x) = `lim_(h->0)` (−2 (−1 + h) −1) = 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 discontinuity.