Question

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?

Answer 1

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(x)` 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

`(x+3)(x-3)=0`

So division by zero occurs when `x=+-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 `0/0` situation.

For this function, it is easy to see that when `x=3`, we get `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]}