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Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain: f(x) =x2+x-3, for x∈[-5,-2)=x2-5, for x∈(-2,5] - Mathematics and Statistics

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प्रश्न

Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:

f(x) `{:(= x^2 + x - 3,","  "for"  x ∈ [ -5, -2)),(= x^2 - 5,","  "for"  x ∈ (-2, 5]):}`

बेरीज
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उत्तर

f is continuous in [– 5, – 2) and in (– 2, 5] since it is a polynomial function.

Continuity at x = – 2

f(x) = x2 + x – 3, for x ∈ [– 5, – 2)

∴ `lim_(x -> - 2^-) "f"(x) = lim_(x -> -2) (x^2 + x - 3)` = 4 – 2 – 3 = – 1

Also, f(x) = x2 + 5, for x ∈ [– 2, 5)

∴ `lim_(x -> -2^+) "f"(x) = lim_(x -> - 2) (x^2 - 5)` = 4 – 5 = – 1

∴ `lim_(x -> -2^-) "f"(x) = lim_(x -> - 2^+) "f"(x)` = – 1

∴ `lim_(x -> - 2) "f"(x)` = – 1

But f(– 2) is not defined.

∴ f is discontinuous at x = – 2

This discontinuity is removable and can be removed by redefining the function as follows:

f(x) `{:(= x^2 + x - 3, ","  "for"  x ∈ [ -5, -2)),(= x^2 - 5, ","  "for"  x ∈ (-2, 5]),(= -1, ","  "for"  x = -2):}`

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Continuous and Discontinuous Functions
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 8 Continuity
MISCELLANEOUS EXERCISE-8 | Q (III) (1) | पृष्ठ १७७

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