Functions 1.2 (Domain)

$f(x) = \sqrt{1-3^{-x^2-2x-3}} =\sqrt{1-\frac 1{3^{x^2+2x+3}}}=\sqrt{1-\frac 1{3^{(x+1)^2+2}}}$

Since $(x+1)^2 \ge 0$ , $\implies 0 < \dfrac 1{3^{(x+1)^2+2}} \le \dfrac 19$ and $\dfrac {2\sqrt 2}3 \le f(x) < 1$ for all real $x$ . Therefore, the domain of $f(x)$ is $\boxed{(-\infty, \infty)}$ .

Let's re-write the equation:

$1-\sqrt{1-3^{-(x^2 + 2x + 3)}}$

$= 1- \sqrt{1 - 3^{-((x+1)^2 +2))}}$

The range of $(x+1)^2 + 2$ is $(2, \infty)$ . That means that the maximum value for $3^{-((x+1)^2 +2))}$ is $3^{-2}$ , which is less than 1.

Therefore, the domain of the function is $\boxed{(-\infty, +\infty)}$ .

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