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