Range of `f(x) =sin^-1(sqrt(x^2+x+1))` is

Answer

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

### Step 1: Analyze the expression under the square root
We start with the expression inside the square root, which is \( x^2 + x + 1 \). This is a quadratic polynomial.

### Step 2: Determine the minimum value of the quadratic
To find the minimum value of \( x^2 + x + 1 \), we can use the vertex formula for a quadratic function \( ax^2 + bx + c \). The vertex \( x \) coordinate is given by \( x = -\frac{b}{2a} \).

Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \):
\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \]

### Step 3: Calculate the minimum value
Now substituting \( x = -\frac{1}{2} \) back into the polynomial:
\[ p\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]
Thus, the minimum value of \( x^2 + x + 1 \) is \( \frac{3}{4} \).

### Step 4: Determine the range of the square root
Since \( x^2 + x + 1 \) has a minimum value of \( \frac{3}{4} \) and goes to infinity as \( x \) increases or decreases, we have:
\[ \sqrt{x^2 + x + 1} \geq \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
and it can take values up to infinity.

### Step 5: Set the bounds for the inverse sine function
The function \( f(x) = \sin^{-1}(\sqrt{x^2 + x + 1}) \) is defined for values of \( \sqrt{x^2 + x + 1} \) between \( \frac{\sqrt{3}}{2} \) and \( 1 \) (since \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\)).

### Step 6: Find the range of \( f(x) \)
Thus, we need to evaluate:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \quad \text{and} \quad \sin^{-1}(1) \]
We know:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \quad \text{and} \quad \sin^{-1}(1) = \frac{\pi}{2} \]

### Conclusion
Therefore, the range of the function \( f(x) \) is:
\[ \left[\frac{\pi}{3}, \frac{\pi}{2}\right] \]