 Find the domain and the range of g(x)=sin^-1(3x+1). 

To find the domain and range of the function \( g(x) = \sin^{-1}(3x + 1) \), we'll analyze the function step by step.

### 1. **Domain:**
The function \( g(x) = \sin^{-1}(y) \) is defined only when the argument \( y \) lies within the interval \([-1, 1]\). Therefore, we require:

\[
-1 \leq 3x + 1 \leq 1
\]

We solve this inequality:

#### Step 1: Subtract 1 from all sides of the inequality:

\[
-1 - 1 \leq 3x + 1 - 1 \leq 1 - 1
\]

\[
-2 \leq 3x \leq 0
\]

#### Step 2: Divide the entire inequality by 3:

\[
-\frac{2}{3} \leq x \leq 0
\]

Thus, the **domain** of \( g(x) \) is:

\[
x \in \left[-\frac{2}{3}, 0\right]
\]

### 2. **Range:**
The range of the inverse sine function \( \sin^{-1}(y) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Since \( g(x) = \sin^{-1}(3x + 1) \), and the expression \( 3x + 1 \) covers the entire interval \([-1, 1]\) within the domain we found, the range of \( g(x) \) remains the same as the range of the inverse sine function.

Thus, the **range** of \( g(x) \) is:

\[
g(x) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\]

### Final Answer:
- **Domain:** \( \left[-\frac{2}{3}, 0\right] \)
- **Range:** \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)

Would you like further details or have any questions?

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**Related Questions:**
1. How does the domain of a function affect its range?
2. What are the steps to find the domain of an inverse trigonometric function?
3. Can you explain why the range of \( \sin^{-1}(y) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)?
4. What would happen if the function was \( g(x) = \cos^{-1}(3x + 1) \)?
5. How does the transformation \( 3x + 1 \) affect the graph of the sine function?

**Tip:** Always ensure the argument of an inverse trigonometric function lies within its principal range to determine the domain correctly.

Find the domain and the range of g(x)=sin^-1(3x+1).

Understand how to find the domain and range of the function g(x) = sin^-1(3x + 1) with a comprehensive step-by-step solution. This problem involves key concepts of inverse trigonometric functions and solving inequalities. Ideal for high school students in Grades 11-12.

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