Question

A function f is defined as follows:

`f(x) = {{:(0,  "for"  x < 0;),(x,  "for"  0 ≤ x ≤ 1;),(- x^2 +4x - 2, "for"  1 ≤ x ≤ 3;),(4 - x,  "for"  x ≥ 3):}`
Is the function continuous?

Answer 1

`f(x) = {{:(0,  "for"  x < 0),(x,  "for"  0 ≤ x ≤ 1),(- x^2 +4x - 2, "for"  1 ≤ x ≤ 3),(4 - x,  "for"  x ≥ 3):}`

`lim_(x -> 0^-) f(x) =  lim_(x -> 0^-) 0` = 0

`lim_(x -> 0^+) f(x) =  lim_(x -> 0^+) x` = 0

∴ `lim_(x -> 0-) f(x) =  lim_(x -> 0^+) f(x)` = 0

Hence `lim_(x -> 0) f(x)` = 0  .......(1)

`f(0)` = 0  .......(2)

From equations (1) and (2) we get

`lim_(x -> 0) f(x) = f(0)`

∴ f(x) is continuos at x = 0.

`lim_(x -> 1^-) f(x) =  lim_(x -> 1^-) x` = 1

`lim_(x -> 1^-) f(x) =  lim_(x -> 1^+) ( x^2 + 4x - 2)`

= – 1² – 4 × 1 – 2

= – 1 + 4 – 2

= 4 – 3

= 1

`lim_(x -> 1^-) f(x)` = 1

∴ `lim_(x -> 1) f(x) =  lim_(x -> 1^+) f(x)` = 1  

Hence `lim_(x -> 1) f(x)` = 1  ........(3)

`f(1) = -1^2 + 4 xx 1 - 2`

= `-1 + 4 - 2`

`f(1)` = 4 – 3

= 1  ........(4)

Fom equaton (3) and (4) we have

`lim_(x -> 1) f(x) = f(1)`

∴ f(x) is continuos at x = 1.

`lim_(x -> 3^-) f(x) =  lim_(x -> 3^-) (- x^2 + 4x - 2)`

= `- 3^2 + 4 xx 3 - 2`

= `- 9 + 12 - 2`

= `- 11 + 12`

= 1

`lim_(x -> 3^-) f(x)` = 1

`lim_(x -> 3^+) f(x) =  lim_(x -> 3^+) (4 - x)`

`lim_(x -> 3^+) f(x)` = 4 – 3

= 1