Find the range of `sin^(-1)sqrt(x^(2)+x+1)`

Explanation

To find the range of the function \( y = \sin^{-1}(\sqrt{x^2 + x + 1}) \), we will follow these steps:

### Step 1: Determine the expression inside the inverse sine function
We start with the expression inside the inverse sine function:
\[ \sqrt{x^2 + x + 1} \]

### Step 2: Analyze the quadratic expression
The expression \( x^2 + x + 1 \) is a quadratic function. To find its minimum value, we can use the vertex formula for a quadratic equation \( ax^2 + bx + c \), where the vertex \( x \) coordinate is given by:
\[ x = -\frac{b}{2a} \]
Here, \( a = 1 \) and \( b = 1 \), so:
\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]

### Step 3: Calculate the minimum value of the quadratic
Now, we substitute \( x = -\frac{1}{2} \) back into the quadratic to find the minimum value:
\[ x^2 + x + 1 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 4: Determine the minimum value of the square root
Since \( \sqrt{x^2 + x + 1} \) is an increasing function, its minimum value occurs at the minimum value of \( x^2 + x + 1 \):
\[ \text{Minimum of } \sqrt{x^2 + x + 1} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 5: Determine the maximum value of the square root
As \( x \) approaches infinity, \( x^2 + x + 1 \) also approaches infinity, hence:
\[ \text{Maximum of } \sqrt{x^2 + x + 1} = \infty \]

### Step 6: Establish the range of the square root function
Thus, the range of \( \sqrt{x^2 + x + 1} \) is:
\[ \left[\frac{\sqrt{3}}{2}, \infty\right) \]

### Step 7: Determine the range of the inverse sine function
The function \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\) and is an increasing function. Therefore, we need to restrict the range of \( \sqrt{x^2 + x + 1} \) to the interval \([0, 1]\):
- The minimum value of \( \sqrt{x^2 + x + 1} \) is \( \frac{\sqrt{3}}{2} \approx 0.866 \), which is greater than \( 0.5 \).
- The maximum value approaches infinity, but we only consider values up to \( 1 \).

Thus, the effective range of \( \sin^{-1}(\sqrt{x^2 + x + 1}) \) is:
\[ \left[\sin^{-1}\left(\frac{\sqrt{3}}{2}\right), \sin^{-1}(1)\right] \]

### Step 8: Calculate the specific values
Calculating these values:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}, \quad \sin^{-1}(1) = \frac{\pi}{2} \]

### Final Answer
Therefore, the range of \( \sin^{-1}(\sqrt{x^2 + x + 1}) \) is:
\[ \left[\frac{\pi}{3}, \frac{\pi}{2}\right] \]