Step 1: Understanding the Concept:
The function is composed of an inverse sine function and a quadratic expression under a square root. We must first ensure the argument of \( \sin^{-1} \) is within its domain \( [-1, 1] \), and then determine the range of valid outputs.
Step 2: Key Formula or Approach:
1. Domain of \( \sin^{-1}(u) \) is \( u \in [-1, 1] \).
2. Minimum value of quadratic \( ax^2+bx+c \) (where \( a\textgreater0 \)) occurs at \( x = -b/2a \).
3. Range of \( \sin^{-1}u \) is increasing with \( u \).
Step 3: Detailed Explanation:
Let \( u = \sqrt{x^2+x+1} \).
For \( f(x) \) to be defined, we need \( -1 \le u \le 1 \). Since \( u \) is a square root, \( u \ge 0 \). Thus, we require:
\[ 0 \le \sqrt{x^2+x+1} \le 1 \]
Squaring the inequality:
\[ 0 \le x^2+x+1 \le 1 \]
First, let's find the minimum value of the expression inside the root, \( g(x) = x^2+x+1 \).
Completing the square:
\[ x^2+x+1 = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4} \]
The minimum value is \( \frac{3}{4} \) at \( x = -\frac{1}{2} \).
So, the minimum value of \( u = \sqrt{x^2+x+1} \) is \( \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).
Combining this with the upper bound required for the domain of \( \sin^{-1} \):
\[ \frac{\sqrt{3}}{2} \le \sqrt{x^2+x+1} \le 1 \]
So, the valid range of the argument for the \( \sin^{-1} \) function is \( \left[ \frac{\sqrt{3}}{2}, 1 \right] \).
Now, we map this interval through \( \sin^{-1} \). Since \( \sin^{-1} \) is strictly increasing:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \le f(x) \le \sin^{-1}(1) \]
Using standard trigonometric values:
\[ \frac{\pi}{3} \le f(x) \le \frac{\pi}{2} \]
Step 4: Final Answer:
The range of the function is \( \left[\frac{\pi}{3}, \frac{\pi}{2}\right] \).