Question

How do you find the x values at which `f(x)=(x+2)/(x^2-3x-10)` is not continuous, which of the discontinuities are removable?

Answer 1

`x=5` is a non removable discontinuity, while `x=-2` can be removed by simplifying the function

Explanation:

As `f(x)` is a rational function the only values where it is not continuous are the values of `x` for which the denominator is null:

`x^2-3x-10 = 0`

`x = frac (3+-sqrt(9+40)) 2 = frac(3+-7) 2`

that is:

`x=-2` and `x=5`

We can therefore factorize the denominator as:

`x^2-3x-10 = (x+2)(x-5)`

and conclude that `x=5` is a non removable discontinuity, while `x=-2` can be removed by simplifying the function and writing it as:

`f(x) = frac (x+2) (x^2-3x-10) = frac (x+2) ((x+2)(x-5)) = 1/(x-5)`