What are the asymptotes and removable discontinuities, if any, of #f(x)=(x^3-x+2)/((x-x^2)(1-x^2))#?
1 Answer
There are none.
Explanation:
Removable discontinuities exist when the function cannot be evaluated at a certain point, but the left and right hand limits equal each other at that point. One such example is the function x/x. This function is clearly 1 (almost) everywhere, but we cannot evaluate it at 0 because 0/0 is undefined. However, the left- and right-hand limits at 0 are both 1, so we can "remove" the discontinuity and give the function a value of 1 at x=0.
When your function is defined by a polynomial fraction, removing discontinuities is synonymous with cancelling factors. If you have time and you know how to differentiate polynomials, I encourage you to prove this for yourself.
Factoring your polynomial is tricky. However, there is an easy way to check where the discontinuities are. First, find all x such that the denominator is 0. To do this, you can factor the denominator as follows:
The first term I factored by pulling out a common factor of x. The second term is the difference of squares,
Here we can see the zeros in the denominator are x=0, x=1, and x=-1.
Without factoring the numerator we can check if the zeros exist in the numerator polynomial. If they do, we will have to do some factoring. If they don't, then we can rest assured that there aren't any factors that would cancel anyway.
In all three cases we got 2, which is not 0. Thus we can conclude that none of the zeros in the denominator match a 0 in the numerator, so none of the discontinuities can be removed.
You can also check this yourself in your graphing software of choice. You will find the function diverges at x = -1, 0, and 1. If the discontinuities were removable, it should look relatively flat in the region around the discontinuity, instead of diverging.
