Find the domain for `f(x)=sin^(-1)((x^(2))/(2))`.

Answer

To find the domain of the function \( f(x) = \sin^{-1}\left(\frac{x^2}{2}\right) \), we need to ensure that the argument of the inverse sine function, \( \frac{x^2}{2} \), lies within the valid range for the sine inverse function. The sine inverse function, \( \sin^{-1}(y) \), is defined for \( y \) in the interval \([-1, 1]\).

### Step-by-step solution:

1. **Set up the inequality**: Since \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\), we need to solve the following inequality:
\[ -1 \leq \frac{x^2}{2} \leq 1 \]

2. **Solve the left side of the inequality**:
\[ -1 \leq \frac{x^2}{2} \]
Since \( \frac{x^2}{2} \) is always non-negative (as \( x^2 \geq 0 \)), this part of the inequality is always satisfied for all real numbers \( x \).

3. **Solve the right side of the inequality**:
\[ \frac{x^2}{2} \leq 1 \]
Multiply both sides by 2:
\[ x^2 \leq 2 \]
Taking the square root of both sides, we get:
\[ -\sqrt{2} \leq x \leq \sqrt{2} \]

4. **Combine the results**: The domain of \( f(x) \) is the set of all \( x \) that satisfy the above inequality. Therefore, the domain is:
\[ x \in [-\sqrt{2}, \sqrt{2}] \]

### Final Answer:
The domain of the function \( f(x) = \sin^{-1}\left(\frac{x^2}{2}\right) \) is:
\[ [-\sqrt{2}, \sqrt{2}] \]