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Find all points of discontinuity of f, where f is defined by: f(x) = {|ЁЭСе| + 3, if ЁЭСе тЙд тИТ3,тИТ2тБвЁЭСе, if тИТ3 < ЁЭСе < 3, 6тБвЁЭСе + 2, if ЁЭСе тЙе 3 - Mathematics

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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):}`

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f(x) = `{(|x|+3", if"  x<= -3),(-2x", if" -3 < x < 3),(6x + 2", if"  x >= 3):}`

⇒ At x = −3

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

= `lim_(h -> 0)` [|−3 − h| + 3]

= `lim_(h -> 0)` (6 + h)

= 6 + 0

= 6

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

= `lim_(h -> 0)` [−2 (−3 + h)]

= `lim_(h -> 0)` (6 − 2h)

= 6 − 2 × 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 + 2h)

= −6 + 2 × 0

= −6

`lim_(x -> 3^+)` f(x) = `lim_(x -> 3^+)` (6x + 2)

= `lim_(h -> 0)` [6(3 + h) + 2]

= `lim_(h -> 0)` (18 + 6h + 2)

= `lim_(h -> 0)` (20 + 6h)

= 20 + 6 × 0

= 20

Hence, f is not continuous at x = 3.

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рдЕрдзреНрдпрд╛рдп 5: Continuity and Differentiability - Exercise 5.1 [рдкреГрд╖реНрда резрелреп]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдЕрдзреНрдпрд╛рдп 5 Continuity and Differentiability
Exercise 5.1 | Q 7 | рдкреГрд╖реНрда резрелреп

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