To find the domain of the function
\[
f(x) = \sin^{-1}\left(\frac{3x^2 + x - 1}{(x - 1)^2}\right) + \cos^{-1}\left(\frac{x - 1}{x + 1}\right),
\]
we need to ensure that the arguments of both the inverse sine and inverse cosine functions are within their respective domains.
### Step 1: Determine the domain for \(\sin^{-1}\)
The argument of \(\sin^{-1}\) must satisfy:
\[
-1 \leq \frac{3x^2 + x - 1}{(x - 1)^2} \leq 1.
\]
#### Part A: Solve \( \frac{3x^2 + x - 1}{(x - 1)^2} \geq -1 \)
This leads to:
\[
3x^2 + x - 1 + (x - 1)^2 \geq 0.
\]
Expanding \((x - 1)^2\):
\[
3x^2 + x - 1 + x^2 - 2x + 1 \geq 0,
\]
\[
4x^2 - x \geq 0.
\]
Factoring gives:
\[
x(4x - 1) \geq 0.
\]
The critical points are \(x = 0\) and \(x = \frac{1}{4}\). Testing intervals, we find:
- For \(x < 0\): \(x(4x - 1) < 0\)
- For \(0 < x < \frac{1}{4}\): \(x(4x - 1) < 0\)
- For \(x > \frac{1}{4}\): \(x(4x - 1) > 0\)
Thus, the solution for this inequality is:
\[
x \in (-\infty, 0] \cup \left[\frac{1}{4}, \infty\right).
\]
#### Part B: Solve \( \frac{3x^2 + x - 1}{(x - 1)^2} \leq 1 \)
This leads to:
\[
3x^2 + x - 1 - (x - 1)^2 \leq 0.
\]
Expanding gives:
\[
3x^2 + x - 1 - (x^2 - 2x + 1) \leq 0,
\]
\[
2x^2 + 3x - 2 \leq 0.
\]
Factoring gives:
\[
(2x + 1)(x - 2) \leq 0.
\]
The critical points are \(x = -\frac{1}{2}\) and \(x = 2\). Testing intervals, we find:
- For \(x < -\frac{1}{2}\): \((2x + 1)(x - 2) > 0\)
- For \(-\frac{1}{2} < x < 2\): \((2x + 1)(x - 2) < 0\)
- For \(x > 2\): \((2x + 1)(x - 2) > 0\)
Thus, the solution for this inequality is:
\[
x \in \left[-\frac{1}{2}, 2\right].
\]
### Step 2: Combine the results
Now, we need to find the intersection of the two sets:
1. From \(\sin^{-1}\): \(x \in (-\infty, 0] \cup \left[\frac{1}{4}, \infty\right)\)
2. From \(\cos^{-1}\): \(x \in \left[-\frac{1}{2}, 2\right]\)
The intersection is:
- From \((- \infty, 0]\) and \([- \frac{1}{2}, 2]\): \(x \in [-\frac{1}{2}, 0]\)
- From \([\frac{1}{4}, \infty)\) and \([- \frac{1}{2}, 2]\): \(x \in [\frac{1}{4}, 2]\)
Thus, the overall domain of \(f(x)\) is:
\[
\text{Domain of } f(x) = [-\frac{1}{2}, 0] \cup [\frac{1}{4}, 2].
\]
### Final Answer
The domain of \(f(x)\) is:
\[
[-\frac{1}{2}, 0] \cup [\frac{1}{4}, 2].
\]
