The given function is:

$f(x) = \sqrt{2 - e^x}$

Step 1: Identify the domain of the square root function

The square root function is only defined for non-negative values, meaning the expression inside the square root must be greater than or equal to zero:

$2 - e^x \geq 0$

Step 2: Solve the inequality

$e^x \leq 2$

Now, take the natural logarithm of both sides:

$x \leq \ln(2)$

The value of $\ln(2)$ is approximately $0.6931$.

Step 3: Domain of the function

Thus, the function is continuous for $x$ values where $x \leq \ln(2)$. This gives the interval:

$(-\infty, \ln(2)]$

So, the function $f(x)$ is continuous on the interval $(- \infty, \ln(2)]$.

Would you like more details or further clarifications on this?

Here are some related questions for deeper understanding:

1. What happens if the value inside the square root becomes negative?
2. How would the domain change if we had $f(x) = \sqrt{3 - e^x}$?
3. Can you determine the range of $f(x)$?
4. How does the function behave as $x$ approaches $\ln(2)$?
5. What would the graph of $f(x)$ look like for this domain?

Tip: Whenever dealing with square roots, always ensure that the radicand (the part inside the root) is non-negative to avoid complex values.