Find the range of the function f(x) `=sin^(-1)sqrt(1+x^4)/(1+5x^10)`

Answer

To find the range of the function \( f(x) = \sin^{-1}\left(\frac{\sqrt{1+x^4}}{1+5x^{10}}\right) \), we will analyze the expression inside the inverse sine function.

### Step 1: Determine the domain of the inner function
The function \( \sin^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to ensure that:
\[ \frac{\sqrt{1+x^4}}{1+5x^{10}} \in [0, 1] \]

### Step 2: Analyze the numerator and denominator
- The numerator \( \sqrt{1+x^4} \) is always non-negative since the square root function is defined for all real numbers, and \( x^4 \) is always non-negative.
- The denominator \( 1 + 5x^{10} \) is also always positive for all real \( x \).

### Step 3: Find the maximum value of the inner function
To find the maximum value of \( \frac{\sqrt{1+x^4}}{1+5x^{10}} \), we can analyze its behavior as \( x \) varies:
1. As \( x \to 0 \):
\[ \frac{\sqrt{1+0^4}}{1+5 \cdot 0^{10}} = \frac{\sqrt{1}}{1} = 1 \]
2. As \( x \to \infty \):
\[ \frac{\sqrt{1+x^4}}{1+5x^{10}} \approx \frac{x^2}{5x^{10}} = \frac{1}{5x^8} \to 0 \]

### Step 4: Find the minimum value of the inner function
Since \( \sqrt{1+x^4} \) is always greater than or equal to 1, we can conclude:
\[ \frac{\sqrt{1+x^4}}{1+5x^{10}} \geq \frac{1}{1+5x^{10}} \to 0 \text{ as } x \to \infty \]
Thus, the minimum value approaches 0 but never actually reaches it.

### Step 5: Determine the range of \( f(x) \)
Since \( \frac{\sqrt{1+x^4}}{1+5x^{10}} \) varies from 0 to 1, the argument of the inverse sine function \( \sin^{-1}(y) \) will vary from:
\[ \sin^{-1}(0) = 0 \quad \text{to} \quad \sin^{-1}(1) = \frac{\pi}{2} \]

### Conclusion
Therefore, the range of the function \( f(x) \) is:
\[ [0, \frac{\pi}{2}] \]