To find the domain of the function $g(t) = \sqrt{3 - t} - \sqrt{2 + t}$, we need to determine the values of $t$ for which the function is defined.

The function involves two square root terms: $\sqrt{3 - t}$ and $\sqrt{2 + t}$. For each square root to be defined, the expression inside the square root must be non-negative. Let's analyze each term separately:

1. Condition for $\sqrt{3 - t}$:

   $3 - t \geq 0$

   Solving for $t$:

   $t \leq 3$

2. Condition for $\sqrt{2 + t}$:

   $2 + t \geq 0$

   Solving for $t$:

   $t \geq -2$

Both conditions must be satisfied simultaneously. Therefore, we need to find the intersection of the intervals:

• $t \leq 3$
• $t \geq -2$

The intersection of these intervals is:

$-2 \leq t \leq 3$

Thus, the domain of the function $g(t)$ is all $t$ values in the closed interval $[-2, 3]$.

Conclusion

The domain of $g(t) = \sqrt{3 - t} - \sqrt{2 + t}$ is:

$\boxed{[-2, 3]}$

Would you like more details on the steps, or do you have any other questions?

Here are 5 related questions to expand your understanding:

1. What is the range of the function $g(t) = \sqrt{3 - t} - \sqrt{2 + t}$ over its domain?
2. How would the domain change if the function were $h(t) = \sqrt{3 - t} + \sqrt{2 + t}$?
3. How can you graph the function $g(t) = \sqrt{3 - t} - \sqrt{2 + t}$?
4. What are the points where $g(t) = 0$ within the domain?
5. How would you find the derivative of $g(t) = \sqrt{3 - t} - \sqrt{2 + t}$?

Tip: When determining the domain of functions involving square roots, always set the expression inside the square root greater than or equal to zero to ensure real-valued outputs.