The given function is:

$f(x) = \sqrt{\frac{8x}{x + 1}}$

To determine where this function is continuous, we need to check:

1. Domain of the square root: The expression inside the square root must be non-negative. Therefore, we need:

$\frac{8x}{x + 1} \geq 0$

2. Denominator must not be zero: The denominator of the fraction $x + 1$ should not be zero, so $x \neq -1$.

Step 1: Solve $\frac{8x}{x + 1} \geq 0$

We need to find the values of $x$ where the fraction is non-negative.

To do this, analyze the critical points of the expression:

• The numerator, $8x$, is zero at $x = 0$.
• The denominator, $x + 1$, is zero at $x = -1$.

Now, let's consider the sign of the fraction $\frac{8x}{x + 1}$ in the following intervals:

• Interval 1: $x < -1$
• Interval 2: $-1 < x < 0$
• Interval 3: $x > 0$

Sign analysis:

• For $x < -1$, both $8x$ and $x + 1$ are negative, so the fraction is positive.
• For $-1 < x < 0$, $8x$ is negative, and $x + 1$ is positive, so the fraction is negative.
• For $x > 0$, both $8x$ and $x + 1$ are positive, so the fraction is positive.

Thus, $\frac{8x}{x + 1} \geq 0$ holds in the intervals $(-\infty, -1) \cup (0, \infty)$.

Step 2: Exclude points where the denominator is zero

We exclude $x = -1$, as the denominator is zero there. Therefore, the function is undefined at $x = -1$.

Step 3: Conclusion on Continuity

The function is continuous where the expression $\frac{8x}{x + 1} \geq 0$, except where the denominator is zero at $x = -1$.

Thus, the function is continuous for $x \in (-\infty, -1) \cup (0, \infty)$.

Answer:

The function is continuous for all values of x except $x = -1$.

Which option corresponds to this?

• The answer is not explicitly listed, but based on the function's analysis, the function is continuous for all $x$ except $x = -1$ (and it must be greater than or equal to 0 for the square root). However, the choice that best matches in spirit (even if not phrased exactly the same) would be G: "The function is not continuous for all values of x less than 0 and greater than or equal to -1."

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Would you like me to clarify anything further or provide additional explanations?

Here are 5 follow-up questions:

1. What would happen if the fraction inside the square root were negative?
2. How does the continuity of rational functions generally depend on the domain of the function?
3. Can you think of any real-world applications where this function might model something continuous?
4. How would the solution change if the square root was removed?
5. What role does the concept of limits play in determining continuity at specific points?

Tip: Always remember that functions involving square roots must have non-negative values under the root for real numbers to be considered valid.