The domain of f(x) = frac{{log_2(x + 3)}}{{x^ | vedclass

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(D) For the function $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ to be defined:$1$. The argument of the logarithm must be positive: $x + 3 > 0 \implie

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$(-3, +\infty) - \{-1, -2\}$

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(D) For the function $f(x) = \frac{\log_2(x + 3)}{x^2 + 3x + 2}$ to be defined:$1$. The argument of the logarithm must be positive: $x + 3 > 0 \implies x > -3$.$2$. The denominator must not be zero: $x^2 + 3x + 2 
eq 0$.Factorizing the denominator: $(x + 1)(x + 2) 
eq 0$,which implies $x 
eq -1$ and $x 
eq -2$.Combining these conditions,the domain is $x \in (-3, \infty)$ excluding the points $\{-1, -2\}$.Thus,the domain is $(-3, \infty) - \{-1, -2\}$.

The domain of $f(x) = \frac{{\log_2(x + 3)}}{{x^2 + 3x + 2}}$ is

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