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Question
Find the points of discontinuity of the function f, where `f(x) = {{:(x^3 - 3",", "if" x ≤ 2),(x^2 + 1",", "if" x < 2):}`
Solution
Clearly, the given function is defined at all points of R.
Case (i) At x = 2
`lim_(x -> 2^-) f(x) = lim_(x -> 2^-) (x^3 - 3)`
= 23 – 3
`lim_(x -> 2^-) f(x)` = 8 – 3 = 5
`lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (x^2 + 1)`
= 22 + 1
`lim_(x -> 2^+) f(x)` = 4 + 1 = 5
∴ `lim_(x -> 2^-) f(x) = lim_(x -> 2^+) f(x)` = 5
Hence `lim_(x -> 2) f(x)` exists.
`f(2)` = 23 – 3
= 8 – 3
= 5
`lim_(x -> 2) f(x) = f(2)`
∴ `f(x)` is continuous at x = 2.
Case (ii) For x < 2
Let x0 be an arbitrary point in `(- oo, 2)`.
`lim_(x -> x_0) f(x) = lim_(x -> x_0) x^3 - 3`
= `x_0^3 - 3`
`f(x_0) = x_0^3 -3`
∴ `lim_(x -> x_0) f(x) = f(x_0)`
∴ f(x)is continuous at x = x0 in `(- oo, 2)`.
Since x0 is an arbitrary point in `(- oo, 2)`.
f(x) is continuous at all points. f`(- oo, 2)`.
Case (iii) For x > 2
Let y0 be an arbitrary point in (2, ∞).
Then `lim_(x -> y_0) f(x) = im_(x -> y_0) (x^2 + 1)`
= `y_0^2 + 1`
`f(y_0) = y_0^2 + 1`
∴ `lim_(x -> y_0) f(x) = f(y_0)`
Hence, f(x) is continuous at x = y0 in `(2, oo)`.
Since yo is an arbitrary point of `(2, oo)`, f(x) is continuous at all points of `(2, oo)`.
∴ By case (i) case (ii) and case (iii) f(x) is continuous at all points of R.
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