To find \( (f-g)(x) \), we start with the definitions of \( f(x) \) and \( g(x) \):
\[ f(x) = -x^4 + 3x^2 - x + 1 \] \[ g(x) = \sqrt{-3x + 1} - 5 \]
Now we can simply calculate \( (f-g)(x) \):
\[ (f-g)(x) = f(x) - g(x) = -x^4 + 3x^2 - x + 1 - \left( \sqrt{-3x + 1} - 5 \right) \] \[ = -x^4 + 3x^2 - x + 1 - \sqrt{-3x + 1} + 5 \] \[ = -x^4 + 3x^2 - x + 6 - \sqrt{-3x + 1} \]
Thus,
\[ (f-g)(x) = -x^4 + 3x^2 - x + 6 - \sqrt{-3x + 1} \]
Next, we need to consider the domain of \( (f-g)(x) \). The term \( \sqrt{-3x + 1} \) requires its radicand to be non-negative:
\[ -3x + 1 \geq 0 \] \[ 1 \geq 3x \] \[ \frac{1}{3} \geq x \]
or
\[ x \leq \frac{1}{3} \]
So the domain of \( (f-g)(x) \) is \( (-\infty, \frac{1}{3}] \).
Based on this, the correct response is:
\[ (f-g)(x) = -x^4 + 3x^2 - x + 6 - \sqrt{-3x + 1}, \quad \text{and the domain of } (f-g)(x) \text{ is } (-\infty, \frac{1}{3}]. \]