Find all values of 
x
 for which the function is continuous. (Enter your answer in interval notation.)
g(x) = 

x under square root
x3 − 3x2 − 10x
 


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!

Find all values of x for which the function is continuous. (Enter your answer in interval notation.) g(x) = √(x^3 − 3x^2 − 10x)

Learn how to find the values of x for which the function g(x) = √(x^3 − 3x^2 − 10x) is continuous. This step-by-step solution involves solving inequalities, factoring cubic expressions, and using interval notation to describe the domain where the function is continuous. Suitable for Grades 11-12 students studying calculus or advanced algebra.

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