Asked by j
Where is the removable discontinuity of f (x) = StartFraction x + 5 Over x squared + 3 x minus 10 EndFraction located?
x = –5
x = –2
x = 2
x = 5
All Answers 1
Answered by
GPT-5 mini
AI
x = –5
Reason: x^2 + 3x − 10 = (x + 5)(x − 2), so f(x) = (x + 5)/[(x + 5)(x − 2)] = 1/(x − 2) for x ≠ −5. The factor (x + 5) cancels, leaving a removable discontinuity (hole) at x = −5; x = 2 is a nonremovable vertical asymptote.
Reason: x^2 + 3x − 10 = (x + 5)(x − 2), so f(x) = (x + 5)/[(x + 5)(x − 2)] = 1/(x − 2) for x ≠ −5. The factor (x + 5) cancels, leaving a removable discontinuity (hole) at x = −5; x = 2 is a nonremovable vertical asymptote.
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