Question

How do you find the domain and range for `f(x)= (x-3)/( x^2- 9)`?

Answer 1

`(x-3)/(x^2-9)`

`=(x-3)/((x-3)(x+3)), x!=3, x!=-3`

`=1/(x+3), x!=3, x!=-3`

The domain will be `RR` \ `{-3,3}`

Remember that `x!=3`, there `1/(3+3)=1/6` is not valid

therefore the range will be `RR`\ `{1/6}`