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Question
Find all points of discontinuity of f, where f is defined by `f(x) = {(x^3 - 3, if x <= 2),(x^2 + 1, if x > 2):}`
Solution
`f(x) = {(x^3 - 3, if x <= 2),(x^2 + 1, if x > 2):}`
For x < 2, f(x) = x3 - 3 and
x > 2, f(x) = x2 + 1 is a polynomial function.
So this is a function.
At x = 2,
`lim_(x -> 2^-)` f(x) = `lim_(x -> 2^-)` (x3 - 3)
`= lim_(h -> 0) [(2 - h)^3 - 3]`
`= lim_(h -> 0) [8 - h^3 - 12 h + 6h^2 - 3]`
`= lim_(h -> 0) (5 - h^3 - 12h + 6h^2) = 5`
`lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (x^2 + 1)`
`= lim_(h -> 0) [(2 + h)^2 + 1]`
`= lim_(h -> 0) (4 + h^2 + 4h + 1)`
`= lim_(h -> 0) (5 + h^2 + 4h)`
= 5
f(2) = (2)3 - 3
= 8 - 3
= 5
Hence, f is a function at x = 2.
There are no points of discontinuity here.
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