f(x) = log₂(x+3)/x²+3x+2 is the domain of

The correct option is (-3, + ∞) – {-1, -2}.

To find the domain of the function given by:

$$f\left(x\right)=\frac{{\log }_2(x+3)}{x^2+3x+2}$$

we need to determine the set of all real numbers $x$ for which the function is defined. This requires satisfying two conditions simultaneously:

Condition 1: The logarithmic term in the numerator must be defined.
The argument of the logarithm must be strictly greater than zero:
$$x+3>0$$ Solving for $x$, we get:
$$x>-3$$ In interval notation, this condition is represented as:
$$x\in (-3,+\infty )$$

Condition 2: The denominator must not equal zero.
To prevent division by zero, the quadratic expression in the denominator must be non-zero:
$$x^2+3x+2\ne 0$$ We factor the quadratic expression:
$$(x+1)(x+2)\ne 0$$ This gives us two restrictions:
$$x\ne -1\text{ and }x\ne -2$$

Combining the Conditions:
We combine the interval from Condition 1 with the exclusions from Condition 2. Both values to exclude, $-1$ and $-2$, lie within the interval $(-3,+\infty )$. Thus, we must remove these points from our domain.

Subtracting the set $\{-1,-2\}$ from $(-3,+\infty )$, we obtain the final domain of the function:
$$\text{Domain}=(-3,+\infty )–\{-1,-2\}$$