To determine where the function $g(x) = \sqrt{x^3 - 3x^2 - 10x}$ is continuous, we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the real number system.

Thus, we require: $x^3 - 3x^2 - 10x \geq 0$ Now, let’s factor the cubic expression: $x^3 - 3x^2 - 10x = x(x^2 - 3x - 10)$ Next, we factor $x^2 - 3x - 10$: $x^2 - 3x - 10 = (x - 5)(x + 2)$ So, the function becomes: $g(x) = \sqrt{x(x - 5)(x + 2)}$ Now, we need to solve the inequality: $x(x - 5)(x + 2) \geq 0$

Step 1: Find the critical points

The critical points occur where each factor equals zero:

• $x = 0$
• $x = 5$
• $x = -2$

Step 2: Test intervals

We will test the sign of $x(x - 5)(x + 2)$ in the following intervals:

1. $(-\infty, -2)$
2. $(-2, 0)$
3. $(0, 5)$
4. $(5, \infty)$

We check the sign of the product in each of these intervals by picking test points within each interval:

1. For $x = -3$ in the interval $(-\infty, -2)$: $(-3)(-3 - 5)(-3 + 2) = (-3)(-8)(-1) = -24 \quad \text{(negative)}$

2. For $x = -1$ in the interval $(-2, 0)$: $(-1)(-1 - 5)(-1 + 2) = (-1)(-6)(1) = 6 \quad \text{(positive)}$

3. For $x = 1$ in the interval $(0, 5)$: $(1)(1 - 5)(1 + 2) = (1)(-4)(3) = -12 \quad \text{(negative)}$

4. For $x = 6$ in the interval $(5, \infty)$: $(6)(6 - 5)(6 + 2) = (6)(1)(8) = 48 \quad \text{(positive)}$

Step 3: Analyze the solution

The expression $x(x - 5)(x + 2) \geq 0$ is satisfied when the product is positive or zero. From our sign test:

• The product is non-negative in the intervals $[-2, 0]$ and $[5, \infty)$.
• The product is zero at $x = -2, 0, 5$.

Thus, the function $g(x)$ is continuous for $x \in [-2, 0] \cup [5, \infty)$.

Final Answer:

The values of $x$ for which the function is continuous are: $\boxed{[-2, 0] \cup [5, \infty)}$

Would you like further details or have any other questions on this?

Here are some related questions for you to explore:

1. How would the continuity change if the function was $\sqrt{x^3 - 3x^2 + 10x}$?
2. What is the domain of the function $f(x) = \sqrt{x^2 + 4x + 3}$?
3. How do you find the points where a piecewise function is continuous?
4. What is the difference between continuity and differentiability?
5. Can a function be continuous at a point but not differentiable there?

Tip: Always check the domain restrictions for square roots—ensure the expression under the root is non-negative!