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):}`

Answer 1

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 y_o 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.