The domain of the real-valued function f(x) = | vedclass

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

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$(-3, -2) \cup (-1, \infty)$

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(D) For the function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ to be defined:$1$. The argument of the logarithm must be positive: $x+3 > 0 \implies x > -3$.$2$. The expression inside the square root in the denominator must be strictly positive: $x^2+3x+2 > 0$.Factoring the quadratic: $(x+2)(x+1) > 0$.This inequality holds when $x \in (-\infty, -2) \cup (-1, \infty)$.Combining the conditions $x > -3$ and $x \in (-\infty, -2) \cup (-1, \infty)$,we get the intersection:$x \in (-3, -2) \cup (-1, \infty)$.

The domain of the real-valued function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ is

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