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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!

Let f(x) = sqrt(8 - e^x). Give the interval(s) on which the function is continuous.

Discover how to find the continuous intervals of the function f(x) = sqrt(8 - e^x). This step-by-step solution explores the domain of square root functions and the behavior of exponential functions, suitable for advanced high school students (Grade 11-12).

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