Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$$ is

We need to find the domain of $$f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$$.

For $$\cos^{-1}(u)$$ to be defined, we need $$-1 \leq u \leq 1$$.

Let $$g(x) = \frac{x^2 - 4x + 2}{x^2 + 3}$$. Note that $$x^2 + 3 > 0$$ for all $$x$$, so the denominator is always positive.

Condition 1: $$g(x) \leq 1$$

$$\frac{x^2 - 4x + 2}{x^2 + 3} \leq 1$$

$$x^2 - 4x + 2 \leq x^2 + 3$$

$$-4x \leq 1$$

$$x \geq -\frac{1}{4}$$

Condition 2: $$g(x) \geq -1$$

$$\frac{x^2 - 4x + 2}{x^2 + 3} \geq -1$$

$$x^2 - 4x + 2 \geq -(x^2 + 3)$$

$$2x^2 - 4x + 5 \geq 0$$

The discriminant is $$16 - 40 = -24 < 0$$, and the coefficient of $$x^2$$ is positive. So this inequality holds for all $$x \in \mathbb{R}$$.

Combining both conditions, the domain is $$\left[-\frac{1}{4}, \infty\right)$$.

The answer is Option B.