The domain of the function $$f(x) = \sin^{-1}\left(\frac{x^2-3x+2}{x^2+2x+7}\right)$$ is

We need to find the domain of $$f(x) = \sin^{-1}\left(\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)$$. For $$\sin^{-1}(u)$$ to be defined, we need $$-1 \leq u \leq 1$$, so we require $$-1 \leq \dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1$$.

We first note that the denominator $$x^2 + 2x + 7 = (x+1)^2 + 6 > 0$$ for all real $$x$$, so we can multiply inequalities without flipping signs.

Condition 1: $$\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1$$. This gives $$x^2 - 3x + 2 \leq x^2 + 2x + 7$$, so $$-5x \leq 5$$, meaning $$x \geq -1$$.

Condition 2: $$\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1$$. This gives $$x^2 - 3x + 2 \geq -(x^2 + 2x + 7) = -x^2 - 2x - 7$$, so $$2x^2 - x + 9 \geq 0$$. The discriminant is $$1 - 72 = -71 < 0$$, and since the leading coefficient is positive, this quadratic is always positive. So this condition is satisfied for all real $$x$$.

Combining both conditions, the domain is $$x \geq -1$$, i.e., $$[-1, \infty)$$.

Hence, the correct answer is Option C: $$[-1, \infty)$$.