To find the domain and range of the function \( f(x) = \sin^{-1}x + \tan^{-1}x + \sec^{-1}x \), we will analyze each component of the function step by step.
### Step 1: Determine the Domain of Each Function Component
1. **Domain of \( \sin^{-1}x \)**:
- The domain of \( \sin^{-1}x \) is \( [-1, 1] \).
2. **Domain of \( \tan^{-1}x \)**:
- The domain of \( \tan^{-1}x \) is \( (-\infty, \infty) \) (all real numbers).
3. **Domain of \( \sec^{-1}x \)**:
- The domain of \( \sec^{-1}x \) is \( (-\infty, -1] \cup [1, \infty) \).
### Step 2: Find the Common Domain
To find the domain of \( f(x) \), we need the intersection of the domains of all three functions:
- \( \sin^{-1}x: [-1, 1] \)
- \( \tan^{-1}x: (-\infty, \infty) \)
- \( \sec^{-1}x: (-\infty, -1] \cup [1, \infty) \)
The common points in these intervals are:
- The only points that fall within \( [-1, 1] \) and also satisfy \( \sec^{-1}x \) are \( -1 \) and \( 1 \).
Thus, the domain of \( f(x) \) is:
\[
\text{Domain} = \{-1, 1\}
\]
### Step 3: Evaluate \( f(x) \) at the Domain Points
1. **Calculate \( f(-1) \)**:
\[
f(-1) = \sin^{-1}(-1) + \tan^{-1}(-1) + \sec^{-1}(-1)
\]
- \( \sin^{-1}(-1) = -\frac{\pi}{2} \)
- \( \tan^{-1}(-1) = -\frac{\pi}{4} \)
- \( \sec^{-1}(-1) = \pi \)
Therefore,
\[
f(-1) = -\frac{\pi}{2} - \frac{\pi}{4} + \pi = \frac{\pi}{4}
\]
2. **Calculate \( f(1) \)**:
\[
f(1) = \sin^{-1}(1) + \tan^{-1}(1) + \sec^{-1}(1)
\]
- \( \sin^{-1}(1) = \frac{\pi}{2} \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)
- \( \sec^{-1}(1) = 0 \)
Therefore,
\[
f(1) = \frac{\pi}{2} + \frac{\pi}{4} + 0 = \frac{3\pi}{4}
\]
### Step 4: Determine the Range
The values obtained from evaluating \( f(x) \) at the points in the domain are:
- \( f(-1) = \frac{\pi}{4} \)
- \( f(1) = \frac{3\pi}{4} \)
Thus, the range of \( f(x) \) is:
\[
\text{Range} = \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\}
\]
### Final Answer
- **Domain**: \( \{-1, 1\} \)
- **Range**: \( \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\} \)
