Find the domain and range of
`f(x)=sin^(-1)x+tan^(-1)x+sec^(-1)x.`

Explanation

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 separately.

### Step 1: Determine the domain of each function component

1. **Domain of \( \sin^{-1}x \)**:
- The domain of \( \sin^{-1}x \) is \( x \in [-1, 1] \).

2. **Domain of \( \tan^{-1}x \)**:
- The domain of \( \tan^{-1}x \) is \( x \in (-\infty, \infty) \) (all real numbers).

3. **Domain of \( \sec^{-1}x \)**:
- The domain of \( \sec^{-1}x \) is \( x \in (-\infty, -1] \cup [1, \infty) \).

### Step 2: Find the overall domain of \( f(x) \)

To find the overall domain of \( f(x) \), we take the intersection of the domains of all three functions:

- The intersection of \( [-1, 1] \) (from \( \sin^{-1}x \)), \( (-\infty, \infty) \) (from \( \tan^{-1}x \)), and \( (-\infty, -1] \cup [1, \infty) \) (from \( \sec^{-1}x \)) gives us:

\[ \text{Domain of } f(x) = [-1, -1] \cup [1, 1] = \{-1, 1\} \]

### Step 3: Evaluate \( f(x) \) at the endpoints of the domain

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 of \( f(x) \)

The values of \( f(x) \) at the endpoints of the domain are:
- \( f(-1) = \frac{\pi}{4} \)
- \( f(1) = \frac{3\pi}{4} \)

Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \]

### Final Answer
- **Domain**: \(\{-1, 1\}\)
- **Range**: \(\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]\)