We are given the function $f(x) = \sqrt{8 - e^x}$. The problem asks to find the interval(s) on which this function is continuous.

Step-by-Step Solution:

1. Square root function domain: The square root function is only defined when the argument inside is non-negative. So, we must have: $8 - e^x \geq 0.$

2. Solve the inequality: $e^x \leq 8.$ Taking the natural logarithm (ln) on both sides, we get: $x \leq \ln(8).$

   Using the approximate value of $\ln(8)$, we get: $\ln(8) \approx 2.079.$

3. Conclusion: Therefore, the function is continuous on the interval $(-\infty, \ln(8)]$, or approximately $(-\infty, 2.079]$.

Final Answer:

The function is continuous on the interval $(-\infty, \ln(8)]$.

Would you like more details on this solution?

Here are some related questions for further practice:

1. How do you find the domain of a composite function like $f(g(x))$?
2. What is the behavior of the exponential function $e^x$ as $x \to \infty$ and $x \to -\infty$?
3. How does the natural logarithm function $\ln(x)$ behave for positive and negative values of $x$?
4. Can you find the range of the function $f(x) = \sqrt{8 - e^x}$?
5. What is the derivative of $f(x) = \sqrt{8 - e^x}$, and how does it affect the function's monotonicity?

Tip: Always check the domain restrictions when dealing with square roots or logarithmic functions to avoid undefined values!