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Mth 271
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Mr. Smith,
I can't find more information on the concept of removable discontinuity. I understand it in principle... You need to define or redefine the function for some values of x, to remove discontinuity... However I am not sure how to use the idea in problems like this one:
f(x) = (x^3 - 5x^2 +6x)/(x^2 + 7x)
Values of x need to be found for which the function has removable discontinuity.
Thank you,
Maja
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@& This function is not defined when x = 0, because the denominator x^2 + 7 x would be zero. So there is a discontinuity at x = 0. There is also a disconintuity at x = -7, the denominator being zero at that point as well.
However we can simplify the equation by first factoring the numerator and the denominator to get
x ( x^2 - 5 x + 6) / ( x ( x + 7) ).
The factor x in the numerator is divided by the factor x in the denominator to give us
(x^2 - 5 x + 6) / (x + 7).
This function has values identical to the original. However whereas the original function had no value at 0, due to being undefined at that point, the current function is perfectly well-behaved at x = 0, taking value 6/7.
So the new function is not only identical to the old one at all its points of definition, it also has a value at 0 which is continuous with those values.
In this sense, then, the discontinuity at x = 0 has been removed.
The new function can be factored further, to the form
(x - 3) ( x - 2) / (x + 7),
but the discontinuity at 7 will remain.*@