# How to Accounting for Removable Discontinuities

Discontinuities at vertical asymptotes can’t be removed. But rational functions sometimes have removable discontinuities in other places. The removable designation is, however, a bit misleading. The gap in the domain still exists at that “removable” spot, but the function values and graph of the curve tend to behave a little better than at x values where there’s a non removable discontinuity. The unction values stay close together — they don’t spread far apart — and the graphs just have tiny holes, not vertical asymptotes where the graphs don’t behave very well (they go infinitely high or infinitely low).

Also Read : The function of a half angle of trigonometric

You have the option of removing discontinuities by factoring the original function statement — if it does factor. If the numerator and denominator don’t have a common factor, then there isn’t a removable discontinuity.

You can recognize removable discontinuities when you see them graphed on a rational function; they appear as holes in the graph — big dots with spaces in the middle rather than all shaded in. Removable discontinuities aren’t big, obvious discontinuities like vertical asymptotes; you have to look carefully for them.

###### Removal by factoring

Discontinuities are removed when they no longer have an effect on the rational function equation. You know this is the case when you find a factor that’s common to both the numerator and the denominator. You accomplish the removal process by factoring the polynomials in the numerator and denominator of the rational function and then reducing the fraction.

To remove the discontinuity in the rational function $\frac{x^2-4}{x^2-5x-14}$, for example, you first factor the fraction into this form:

Now you reduce the fraction to the new function statement:

$y=\frac{x-2}{x-7}$

By getting rid of the removable discontinuity, you simplify the equation that you’re graphing. It’s easier to graph a line with a little hole in it than deal with an equation that has a fraction — and all the computations involved.

Evaluating the removal restrictions

The function $y=\frac{x^-4}{x^2-5x-14}$,

You factor the denominator, and when you set it equal to zero, you find that the solutions x = –2 and x = 7 don’t appear in the domain of the function. Now what?

Numbers excluded from the domain stay excluded even after you remove the discontinuity. The function still isn’t defined for the two values you find. Therefore, you can conclude that the function behaves differently at each of the discontinuities. When x = –2, the graph of the function has a hole; the curve approaches the value, skips it, and goes on. It behaves in a reasonable fashion: The function values skip over the discontinuity, but the values get really close to it. When x = 7, however, a vertical asymptote appears; the discontinuity doesn’t go away. The function values go haywire at that x value and don’t settle down at all.

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