Question

Find all the point of discontinuities of f(x) = [x] on the interval (− 3, 2).

Answer 1

f(x) = [x], x ∈ (−3, 2)

i.e., f(x) = − 3 for x ∈ (− 3, − 2)

= − 2 for x ∈ [− 2, − 1)

= − 1 for x ∈ [− 1, 0)

= 0 for x ∈ [0, 1)

= 1 for x ∈ [1, 2)

 Consider continuity at x = − 2

`lim_(x -> -2^-) "f"(x) =  lim_(x -> -2) (- 3)` = − 3

`lim_(x -> -2^+) "f"(x) =  lim_(x -> -2) (- 2)` = − 2

∴ `lim_(x -> - 2^-) "f"(x) ≠ lim_(x -> - 2^+) "f"(x)`

∴ `lim_(x -> - 2) "f"(x)` does not exist

∴ f is discontinuous at x = − 2

Similarly, f is discontinuous at other integral values of x

i.e., − 1, 0, 1

Let a ∈ (− 3, − 2)

It is clear that

`lim_(x ->  "a"^-) "f"(x) = lim_(x -> "a"^+) "f"(x)` = f(a) = − 3

 ∴ f is continuous in (− 3, − 2)

Similarly, f is continuous in (− 2, − 1), (− 1, 0), (0, 1), (1, 2)

Hence, f is discontinuous at x = − 2, − 1, 0, 1.