Question

Let `f(x)=(x^3 - 4x)/(x^3 +x^2-6x)`, how do you find all points of discontinuity of f(x)?

Answer 1

Factor the numerator and denominator to identify their zeros to determine the points of discontinuity and their types.

Explanation:

`f(x) = (x^3-4x)/(x^3+x^2-6x) = (x(x-2)(x+2))/(x(x-2)(x+3)) = (x+2)/(x+3)`

with exclusions `x != 0` and `x != 2`.

There are removable discontinuities at `x=0` and `x=2` where both the numerator and the denominator are zero, so `f(x)` is undefined at those points, but the left and right limits agree.

There is a simple pole at `x=-3`, where the denominator is zero and the numerator is non-zero.

`lim_(x->-3+) f(x) = -oo`
`lim_(x->-3-) f(x) = +oo`

Apart from these discontinuities, `f(x)` is well defined and continuous.

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

Answer 2

That function has discontinuities at 0, 2, and -3.

Explanation:

A rational function is continuous on its domain.

So the points of discontinuity for a rational function are the point outside the domain.

`x^3+x^2-6x = 0`

`x(x^2+x-6) = x(x-2)(x+3) = 0`

The points outside the domain are: `0, " "2, " and "-3`

Note

Because this was posted in the topic "Classifying Discontinuities, I should probably add that

`f(x) = (x(x+2)(x-2))/(x(x-2)(x+3)) = (x+2)/(x+3)` for `x != 0, 2`

So the only infinite limit occurs at `x=-3`. There is a non-removable (infinite) discontinuity at `-3`

The discontinuities at `0` and `2` are removable.