To solve this problem, let's analyze the given function and determine the intervals over which $f(x)$ is continuous.

Function:

$f(x) = \sqrt{x} (x^3 - 7)$

Step 1: Conditions for continuity

1. Square root $\sqrt{x}$: The square root function is only defined when $x \geq 0$.
2. Polynomial $x^3 - 7$: Polynomials are continuous for all real numbers.

Thus, the overall function $f(x)$ is continuous when:

• $x \geq 0$ (from $\sqrt{x}$),
• No other constraints exist since $x^3 - 7$ is defined for all $x \geq 0$.

Step 2: Interval of continuity

The function is defined and continuous on the interval: $[0, \infty)$

Final Answer:

$\text{interval: } [0, \infty)$

Let me know if you'd like a detailed breakdown of these steps!

5 Related Questions:

1. Why is the square root function only defined for $x \geq 0$?
2. What are the key rules for determining continuity of a function?
3. How do you find the domain of a composite function?
4. What happens if a function contains a logarithm instead of a square root?
5. How can discontinuities appear in piecewise functions?

Tip:

Always start by analyzing each component of a function to determine restrictions on its domain!