Question

finding discontinuities of a rational function
find the discontinuities of the function.
$f(x) = \frac{x^2 + 12x + 27}{x^2 + 4x + 3}$
there is a removable discontinuity at (
-3 ✔️, 6 ❌).

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 + 12x + 27\): \(x^2 + 12x + 27=(x + 3)(x + 9)\)
Factor \(x^2 + 4x + 3\): \(x^2 + 4x + 3=(x + 1)(x + 3)\)
So \(f(x)=\frac{(x + 3)(x + 9)}{(x + 1)(x + 3)}\)

Step2: Identify common factors

The common factor in numerator and denominator is \((x + 3)\), so we can cancel it (for \(x
eq - 3\)). The value that makes the denominator zero (and is canceled) gives the removable discontinuity. Set \(x+3 = 0\), so \(x=-3\).

Answer:

The removable discontinuity is at \(x = - 3\) (or the point \((-3, \frac{(-3)+9}{(-3)+1})=(-3, - 3)\))