#### Answer

The domain of $g(t)$ is $(-\infty,\infty)$
The range of $g(t)$ is $(1,\infty)$

#### Work Step by Step

$$g(t)=\sqrt{1+3^{-t}}=\sqrt{1+\frac{1}{3^t}}$$
1) Domain:
$g(t)$ is only defined as it satisfies 2 condititons: $3^t\ne0$ and $(1+3^{-t})\ge0$
- First, for $3^t\ne0$: since $3\gt0$, we have $3^t\gt0$ for all $t\in R$. Therefore, there is no value of $t$ which would lead to $3^t=0$.
- Second, for $(1+3^{-t})\ge0$:
As $3^t\gt0$ for all $t\in R$, $\frac{1}{3^t}=3^{-t}\gt0$ for all $t\in R$.
That means $1+3^{-t}\gt1$ for all $t\in R$
Therefore, g(t) is defined on $R$. In other words, the domain of $g(t)$ is $(-\infty,\infty)$
2) Range:
- As proved above, $1+3^{-t}\gt1$ as $t$ in the domain. So $\sqrt{1+3^{-t}}\gt1$
In other words, the range of g(t) is $[-1,1]$
- Second, since $t$ is defined up until $-\infty$, $3^t$ can reach value as low as $0$ (though never equals $0$), meaning $\sqrt{1+\frac{1}{3^t}}$ can range up to $\infty$.
Therefore, the range of $g(t)$ would be $(1,\infty)$.