Question

Let `f(x) = {{:(0",",  "if"  x < 0),(x^2",",  "if"  0 ≤ x ≤ 2),(4",",  "if"  x ≥ 2):}`. Graph the function. Show that f(x) continuous on `(- oo, oo)`

Answer 1

`f(x) = {{:(0",",  "if"  x < 0),(x^2",",  "if"  0 ≤ x ≤ 2),(4",",  "if"  x ≥ 2):}`

Let y = f(x)

x	– 1	0	1	2	3	4	5
f(x)	0	0	1	4	4	4	4

When x < 0

We have y = 0

When 0 ≤ x < 2

We have y = x²

When x ≥ 2

We have y = 4

Case (i) If x < 0

i.e. `(-oo, 0)` then f(x) = 0

Which is clearly continuous in `(-oo, 0)`.

Case (ii) If 0 ≤ x < 2

i.e. (0 , 2)

Let x₀ be an arbitrary point in (0, 2)

`lim_(x -> x_0) f(x) =  lim_(x -> x_0)  x^2 = x_0^2`

`f(x_0) = x_0^2`

∴ `lim_(x -> x_0) f(x) = f(x_0)`

Hence f(x) is continuous at x = x₀.

Since x = x₀ is an arbitrary f(x) is continuous at all points of (0, 2).

Case (iii) x ≥ 2

i.e. `(2, oo)`

f(x) = 4 which is clearly continuous in `(2, oo)`

Case (iv) At x = 2,

`lim_(x -> 2^-) f(x) =  lim_(x -> 2^-) x^2` = 2² = 4

`lim_(x -> 2^-) f(x) =  lim_(x -> 2^-) x^2` = 2² = 4

`lim_(x -> 2^+) f(x) =  lim_(x -> 2^-) f(x)` = 4

`lim_(x -> 2) f(x)` = 4

f(2) = 4

∴ f(x) is continuous at x = 2.

∴ Using case (i) case (ii) case (iii) and case (iv).

We have f(x) is continuous at all points of R.