# 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?

##### 1 Answer

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

Removable ones:

#### Explanation:

Any function

What values of

#x^2-9=0#

which we can factor to get

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

So division by zero occurs when

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

For this function, it is easy to see that when

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