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प्रश्न
Find all points of discontinuity of f, where f is defined by `f(x) = {(|x|+3, if x<= -3),(-2x, if -3 < x < 3),(6x + 2, if x >= 3):}`
उत्तर
`f(x) = {(|x|+3, if x<= -3),(-2x, if -3 < x < 3),(6x + 2, if x >= 3):}`
For x < -3, f(x) = `abs x + 3:`
-3 < x < 3, f(x) = - 2x and
x ≥ 3, f(x) = 6x + 2 is a polynomial function.
So this is a function.
⇒ At x = - 3,
`lim_(x -> 3^-) f(x) = lim_(x -> 3^-) (abs x + 3)`
`= lim_(h -> 0) [abs (-3 - h) + 3]`
`= lim_(h -> 0) (6 + h)`
= 6 + 0
= 6
`lim_(x -> 3^+)` f(x) = `lim_(x -> 3^+)` (-2 x)
`= lim_(h -> 0) [-2 (-3 + h)]`
`= lim_(h -> 0) (6 - 2h)`
`= 6 - 2 xx 0`
= 6
Hence, f is continuous at x = -3.
⇒ At x = 3,
`lim_(x -> 3^-)` f(x) = `lim_(x -> 3^-)` (- 2x)
`= lim_(h -> 0) [-2 (3 - h)]`
= `lim_(h -> 0) (- 6 + 2 h)`
`= -6 + 2 xx 0`
= - 6
`lim_(x -> 3^+)` f(x) = `lim_(x -> 3^+)` (6x + 2)
= `lim_(h -> 0)` [6(3 + h) + 2]
= `lim_(h -> 0)` (18 + 6 h + 2)
= `lim_(h -> 0)` (20 + 6h)
= 20 + 6 × 0
= 20
Hence, f is not continuous at x = 3
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