How do you find the x values at which f(x)=(x-3)/(x^2-9) is not continuous, which of the discontinuities are removable?

Dec 22, 2016

All discontinuities: set the denominator equal to 0 and solve for $x$.

Removable ones: $x$-values that make numerator and denominator 0 simultaneously.

Explanation:

Any function $f \left(x\right)$ will be discontinuous at $x$-values that make the function undefined. Here, that will simply be any $x$-value that creates a "division by zero".

What values of $x$ create division by zero? In this case, it will be any $x$ that satisfies

${x}^{2} - 9 = 0$

which we can factor to get

$\left(x + 3\right) \left(x - 3\right) = 0$

So division by zero occurs when $x = \pm 3$.

To find out if either of these values is a removable discontinuity, we check them one at a time to see if they also make the numerator equal to zero, thus creating a $\frac{0}{0}$ situation.

For this function, it is easy to see that when $x = 3$, we get $\frac{0}{0}$, because $x - 3$ is a factor of both sides of the fraction. Thus, $x = 3$ is a removable discontinuity.

graph{(y-(x-3)/(x^2-9))((x-3)^2+(y-0.167)^2-.02)=0 [-11.5, 8.5, -4.046, 5.95]}