The given function is:
$f(x) = \sqrt{6x^2 - 48}$

Step 1: Identify the Domain of $f(x)$

Since the function contains a square root, we must ensure that the expression inside the square root is non-negative:
$6x^2 - 48 \geq 0$

Step 2: Solve for $x$

$6x^2 \geq 48$

$x^2 \geq 8$

$x \leq -\sqrt{8} \quad \text{or} \quad x \geq \sqrt{8}$

Since the function is only defined when the expression inside the square root is non-negative, the domain is:
$(-\infty, -\sqrt{8}] \cup [\sqrt{8}, \infty)$

Step 3: Determine Continuity

• The function $f(x)$ is continuous wherever it is defined because it consists of a square root function, which is continuous on its domain.
• The only potential points of concern are at $x = \pm\sqrt{8}$, where the function transitions between being defined and undefined.

Step 4: Check Left and Right Continuity at Finite Endpoints

• At $x = -\sqrt{8}$:

  • $f(x)$ is continuous from the right (since it is defined and continuous for $x > -\sqrt{8}$).
  • It is not continuous from the left, since the function is undefined for $x < -\sqrt{8}$.

• At $x = \sqrt{8}$:

  • $f(x)$ is continuous from the left (since it is defined and continuous for $x < \sqrt{8}$).
  • It is not continuous from the right, since the function is undefined for $x > \sqrt{8}$.

Final Answer:

The function is continuous on:
$(-\infty, -\sqrt{8}] \cup [\sqrt{8}, \infty)$

At the finite endpoints:

• $f(x)$ is continuous from the right at $x = -\sqrt{8}$.
• $f(x)$ is continuous from the left at $x = \sqrt{8}$.

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Would you like a step-by-step breakdown of why the function is continuous on this interval?

Here are 5 related questions to explore further:

1. How do you determine the continuity of a function at a point?
2. What are the conditions for a function to be continuous on an interval?
3. How does the presence of a square root affect the domain of a function?
4. What is the difference between continuity from the left and continuity from the right?
5. How would the function's continuity change if it contained an absolute value instead of a square root?

Tip: Always check the domain of a function before analyzing continuity!