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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?
Solution
`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
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