Question

find the dicontinuities of the function.
$f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$.
there is a removable discontinuity at ( , ).

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+12x + 27=(x + 3)(x+9)$. The denominator $x^{2}+4x + 3=(x + 1)(x + 3)$. So $f(x)=\frac{(x + 3)(x + 9)}{(x + 1)(x + 3)}$.

Step2: Simplify the function

Cancel out the common factor $(x + 3)$ (for $x
eq - 3$), and we get $f(x)=\frac{x + 9}{x + 1},x
eq-3$.

Step3: Find the discontinuities

The original function $f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$ is undefined when the denominator $x^{2}+4x + 3 = 0$. Solving $(x + 1)(x + 3)=0$, we get $x=-1$ and $x=-3$. At $x=-1$, the function has an infinite discontinuity since $\lim_{x
ightarrow - 1}\frac{x + 9}{x + 1}=\pm\infty$. At $x=-3$, the original function is undefined, but $\lim_{x
ightarrow - 3}\frac{x^{2}+12x + 27}{x^{2}+4x + 3}=\lim_{x
ightarrow - 3}\frac{x + 9}{x + 1}=\frac{-3 + 9}{-3+1}=\frac{6}{-2}=-3$. So there is a removable discontinuity at $x=-3$. When $x=-3$, the value of the simplified - function (after removing the common factor) is $-3$.

Answer:

$(-3,-3)$