Step 1: Identify all expressions that impose domain restrictions.
The function is $f(x) = \frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}$. Three square root expressions appear: $\sqrt{2-x}$ (numerator), $\sqrt{1+x}$ (numerator), and $\sqrt{x+3}$ (denominator). Each imposes its own restriction on $x$, and we need all three satisfied simultaneously.
Step 2: Condition from $\sqrt{2-x}$.
For the square root to produce a real number, its argument must be non-negative (square roots of negatives are not real). So $2-x \geq 0$, giving $x \leq 2$. This is the interval $(-\infty,2]$.
Step 3: Condition from $\sqrt{1+x}$.
Similarly, $1+x \geq 0$ gives $x \geq -1$. This is the interval $[-1,\infty)$.
Step 4: Condition from $\sqrt{x+3}$ in the denominator.
Since this square root is in the denominator, we need it to be strictly positive (zero denominator is undefined): $x+3 > 0$, so $x > -3$. This is the interval $(-3,\infty)$.
Step 5: Intersect all three conditions.
We need: $x \leq 2$, $x \geq -1$, and $x > -3$ all at once. The binding constraints are $x \geq -1$ and $x \leq 2$ (the third constraint $x > -3$ is automatically satisfied when $x \geq -1$). The intersection is $[-1,2]$.
Step 6: Check the boundary points.
At $x=-1$: $\sqrt{3} + 0$ over $\sqrt{2}$, all defined. At $x=2$: $0+\sqrt{3}$ over $\sqrt{5}$, all defined. Both endpoints are included.
\[ \boxed{[-1,2]} \]