To find the range of the function \( f(x) = \sin^{-1} x + |\sin^{-1} x| + \sin^{-1} |x| \), we can analyze the function based on the properties of the inverse sine function and the absolute value.
### Step 1: Analyze the function for different intervals of \( x \)
1. **For \( x \in [0, 1] \):**
- Here, \( \sin^{-1} x \) is non-negative, so:
\[
f(x) = \sin^{-1} x + |\sin^{-1} x| + \sin^{-1} |x| = \sin^{-1} x + \sin^{-1} x + \sin^{-1} x = 3 \sin^{-1} x
\]
- The range of \( \sin^{-1} x \) for \( x \in [0, 1] \) is from \( 0 \) to \( \frac{\pi}{2} \). Therefore, the range of \( f(x) \) in this interval is:
\[
[0, 3 \cdot \frac{\pi}{2}] = [0, \frac{3\pi}{2}]
\]
2. **For \( x \in [-1, 0) \):**
- Here, \( \sin^{-1} x \) is non-positive, so:
\[
f(x) = \sin^{-1} x + |\sin^{-1} x| + \sin^{-1} |x| = \sin^{-1} x - \sin^{-1} x + \sin^{-1} (-x) = \sin^{-1} (-x)
\]
- Since \( -x \) ranges from \( 0 \) to \( 1 \) as \( x \) ranges from \( -1 \) to \( 0 \), we have:
\[
f(x) = \sin^{-1} (-x) \text{ which ranges from } 0 \text{ to } \frac{\pi}{2}
\]
3. **For \( x < -1 \) or \( x > 1 \):**
- The function \( \sin^{-1} x \) is not defined for \( |x| > 1 \), so we only consider \( x \in [-1, 1] \).
### Step 2: Combine the ranges
From the analysis:
- For \( x \in [0, 1] \), the range of \( f(x) \) is \( [0, \frac{3\pi}{2}] \).
- For \( x \in [-1, 0) \), the range of \( f(x) \) is \( [0, \frac{\pi}{2}] \).
Thus, the overall range of \( f(x) \) is:
\[
[0, \frac{3\pi}{2}]
\]
### Final Answer
The range of \( f(x) \) is \([0, \frac{3\pi}{2}]\).
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