Question

How do you find points of discontinuity in rational functions?

Answer 1

Hint: We take an example to illustrate how to find the points of discontinuity. In order to find the point of discontinuity of any function, we simply equate the denominator with zero.

Complete step-by-step solution:
Points of discontinuity in rational functions refer to those fractions which are undefined or which have their denominators as zero. Once a fraction has its denominator is zero, it becomes undefined and then it appears as a hole or a break in the graph.
Let us take a rational function as an example to illustrate this:
Let us take $f\left( x \right) = \dfrac{{{x^2} + 2x - 8}}{{{x^2} + 5x + 4}}$
Now we factories both the numerator and denominator, to do this we first split the middle terms:
\[ \Rightarrow f\left( x \right) = \dfrac{{{x^2} + 4x - 2x - 8}}{{{x^2} + 4x + x + 4}}\]
Now we group the common factors:
$ \Rightarrow f\left( x \right) = \dfrac{{x\left( {x + 4} \right) - 2\left( {x + 4} \right)}}{{x\left( {x + 4} \right) + 1\left( {x + 4} \right)}}$
Thus we successfully factories the numerator and denominator to get our factors as:
$ \Rightarrow f\left( x \right) = \dfrac{{\left( {x + 4} \right)\left( {x - 2} \right)}}{{\left( {x + 4} \right)\left( {x + 1} \right)}}$
In the above expression, we find that $\left( {x + 4} \right)$ can be canceled out from both the denominator and the numerator, thus it becomes a removable discontinuity while $\left( {x + 1} \right)$ cannot be canceled out or factored any further hence it becomes non-removable discontinuity.
Now, this rational function will only be discontinuous when the denominator is equal to zero. To find the discontinuities, we equate the denominator with zero. Thus, we get:
$x + 4 = 0$ and $x + 1 = 0$
$ \Rightarrow x = - 4$ and $x = - 1$ , thus the given rational functions will be discontinuous at these following points.

Note: Let us take another rational function as an example.
Let $f\left( x \right) = \dfrac{{x + 1}}{{{x^2} - 5x - 6}}$ , now to prove discontinuity again we need to equate the denominator with zero
Thus, we find the factors of the denominator
$ \Rightarrow f\left( x \right) = \dfrac{{x + 1}}{{{x^2} - 6x + x - 6}}$
We do grouping to find the common factors:
$ \Rightarrow f\left( x \right) = \dfrac{{x + 1}}{{\left( {x + 1} \right)\left( {x - 6} \right)}}$
Thus, if we equate the factors in the denominator, we find that the function is discontinuous at $x = - 1$ and $x = 6$