Find the range of each of the following function:
` f(x)=sin^(-1)(sqrt(3x^(2)+1)/(5x^(2)+1))`

Explanation

To find the range of the function \( f(x) = \sin^{-1}\left(\frac{\sqrt{3x^2 + 1}}{5x^2 + 1}\right) \), we will follow these steps:

### Step 1: Analyze the Function Inside the Inverse Sine
The function inside the inverse sine is \( \frac{\sqrt{3x^2 + 1}}{5x^2 + 1} \). We need to determine the range of this expression for all real values of \( x \).

### Step 2: Determine the Limits of the Expression
1. **As \( x \to 0 \)**:
\[ \frac{\sqrt{3(0)^2 + 1}}{5(0)^2 + 1} = \frac{\sqrt{1}}{1} = 1 \]

2. **As \( x \to \infty \)**:
\[ \frac{\sqrt{3x^2 + 1}}{5x^2 + 1} \approx \frac{\sqrt{3x^2}}{5x^2} = \frac{\sqrt{3}}{5} \]

3. **As \( x \to -\infty \)**:
The analysis is similar to \( x \to \infty \):
\[ \frac{\sqrt{3x^2 + 1}}{5x^2 + 1} \approx \frac{\sqrt{3}}{5} \]

### Step 3: Find the Range of the Expression
From the above limits, we see that:
- The expression approaches \( 1 \) as \( x \) approaches \( 0 \).
- The expression approaches \( \frac{\sqrt{3}}{5} \) as \( x \) approaches \( \infty \) or \( -\infty \).

### Step 4: Determine the Valid Range for the Inverse Sine Function
Since the value of \( \frac{\sqrt{3x^2 + 1}}{5x^2 + 1} \) is always positive and lies between \( \frac{\sqrt{3}}{5} \) and \( 1 \), we can conclude that:
\[ \frac{\sqrt{3}}{5} < \frac{\sqrt{3x^2 + 1}}{5x^2 + 1} \leq 1 \]

### Step 5: Apply the Inverse Sine Function
The range of \( \sin^{-1}(y) \) where \( y \) is in the interval \( \left(\frac{\sqrt{3}}{5}, 1\right) \) corresponds to:
- \( \sin^{-1}\left(\frac{\sqrt{3}}{5}\right) \) to \( \sin^{-1}(1) \).

### Step 6: Final Range
Since \( \sin^{-1}(1) = \frac{\pi}{2} \) and \( \sin^{-1}\left(\frac{\sqrt{3}}{5}\right) \) is a calculable value, we can denote the range of \( f(x) \) as:
\[ \left(\sin^{-1}\left(\frac{\sqrt{3}}{5}\right), \frac{\pi}{2}\right) \]

### Conclusion
Thus, the range of the function \( f(x) = \sin^{-1}\left(\frac{\sqrt{3x^2 + 1}}{5x^2 + 1}\right) \) is:
\[ \left(\sin^{-1}\left(\frac{\sqrt{3}}{5}\right), \frac{\pi}{2}\right) \]