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Be very careful when you write equations. In a book for example this expression might have been written like this:
x^2-49
---------
x - 7
When you type that you have to add parentheses to preserve the correct order of operations. So type it as
(x^2-49) / (x-7)
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The original has a discontinuity at x = 7 because that would be division by zero.
However, that factors to
(x-7)(x+7)/(x-7)
which simplifies to x + 7. So, by defining f(7) to be 14 you can remove the discontinuity.
The removable discontinuity is at x = 7.
Be very careful when you write equations. In a book for example this expression might have been written like this:
x^2-49
---------
x - 7
When you type that you have to add parentheses to preserve the correct order of operations. So type it as
(x^2-49) / (x-7)
---
The original has a discontinuity at x = 7 because that would be division by zero.
However, that factors to
(x-7)(x+7)/(x-7)
which simplifies to x + 7. So, by defining f(7) to be 14 you can remove the discontinuity.
The removable discontinuity is at x = 7.
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(x^2 - 49) / (x - 7)
. . . x = 7 is a discontinuity because (x - 7) = 0, and you can't divide by zero.
. . . factoring the numerator ...
(x + 7) * (x - 7) / (x - 7)
. . . x = 7 is removable because (x - 7) / (x - 7) = 1 and cancels out to leave
x + 7
. . . x = 7 is a discontinuity because (x - 7) = 0, and you can't divide by zero.
. . . factoring the numerator ...
(x + 7) * (x - 7) / (x - 7)
. . . x = 7 is removable because (x - 7) / (x - 7) = 1 and cancels out to leave
x + 7
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You must have made a typo because there are no removable discontinuities.
f(x) = x² - 49 / x - 7
f(x) = x² - 7 - 49 / x
f(x) = (x³ - 7x - 49) / x
f(x) = x² - 49 / x - 7
f(x) = x² - 7 - 49 / x
f(x) = (x³ - 7x - 49) / x
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x² – 49 = (x + 7)(x – 7) so f(x) has a removable discontinuity at (7, 14)
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f(x) = [(x-7)(x+7)]/(x-7)
= x+7
f(x) = 0
x+7 = 0
x = -7
= x+7
f(x) = 0
x+7 = 0
x = -7
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x = 7, since you cancel x-7.