find the dicontinuities of the function.\n$f(x)=\frac{x^{2}+12x + 27}{x^{2}+4x + 3}$.\nthere is a removable…

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 $x\neq - 3$), we get $f(x)=\frac{x + 9}{x + 1},x\neq - 3$.

Step3: Identify the removable discontinuity

A removable discontinuity occurs when a factor can be canceled out. The original function is undefined at $x=-3$ (because the original denominator is zero at $x=-3$), but the limit as $x\to - 3$ exists. We find the limit $\lim_{x\to - 3}\frac{x^{2}+12x + 27}{x^{2}+4x + 3}=\lim_{x\to - 3}\frac{(x + 3)(x + 9)}{(x + 1)(x + 3)}=\lim_{x\to - 3}\frac{x + 9}{x + 1}=\frac{-3 + 9}{-3+1}=\frac{6}{-2}=-3$.

Answer:

$(-3,-3)$