How to show if a function is continuous

Question

I am faced with this function: $$ h(x)= \frac{1−2x}{x^2−x−6} $$ And I have to determine its points of discontinuity. I am not quite sure how to approach this. Any help?

Thank you!

Answer 1

This depends on the sophistication of the techniques you need to use. Clearly, the rational function $h(x)$ is not continuous at the points of real numbers when the denominator is zero. That is: If $x^2 - x - 6 = 0$, then $h(x)$ has points of discontinuity $(3,0)$ and $(-2,0)$ because

$$\begin{aligned} (x - 3)(x + 2) &= 0\\ x &= 3, -2 \end{aligned}$$

If a function is continuous, then we can find $(x,h(x))$ on the graph of $h(x)$. For instance, the function $h(x)$ at $x = 0$ exists since the denominator does not vanish i.e. the denominator is not zero.

---

Otherwise, if you want to show that the rational function is continuous, then see $h(x)$ as $f(x)/g(x)$; if $g(x)$ is not zero at the point $x$, then $f(x)/g(x)$ is continuous. Note the following:

If $f(x)$ is continuous at $x = a$ and $g(x)$ is continuous at $x = a$, such that $g(a) \neq 0$, then $f(a)/g(a)$ exists, so $f(x)/g(x)$ is continuous at $x = a$.

Answer 2

The facts you should know to solve this:

1. If $f$ and $g$ are both well-defined and continuous on the same interval and $g\neq 0$ on the interval, then so is $f/g$ (i.e. well-defined and continuous).

2. If $g=0$ at some point, then $f/g$ is not defined at the point.

Answer 3

Use graphing software to see what the graph looks like. This isn't a graph with just a missing point or two, when X is 3 or -2 the graph shows an asymptote, which approaches +/- infinity based on which side it approaches the asymptote line.