The domain of the function f (x) = ^ - 1 ( x ^ 2 - 3x + 2 x ^ 2 +...

Given,

$f\left(x\right) = \sin^{- 1}\left(\dfrac{{x}^{2} - 3x + 2}{{x}^{2} + 2x + 7}\right)$

We know that domain of $\sin^{- 1}\left(f\left(x\right)\right)$ is $- 1 \le f\left(x\right) \le 1$

So, $\dfrac{{x}^{2} - 3x + 2}{{x}^{2} + 2x + 7} \ge - 1$ and $\dfrac{{x}^{2} - 3x + 2}{{x}^{2} + 2x + 7} \le 1$

Now solving $\dfrac{{x}^{2} - 3x + 2}{{x}^{2} + 2x + 7} \ge - 1$

Since denominator ${x}^{2} + 2x + 7$ is always positive,

So, ${x}^{2} - 3x + 2 \ge - \left({x}^{2} + 2x + 7\right)$

$⇒ 2{x}^{2} - x + 9 \ge 0$

$⇒ x \in R . . . . . \left(1\right)$

Now solving $\dfrac{{x}^{2} - 3x + 2}{{x}^{2} + 2x + 7} \le 1$

$⇒ {x}^{2} - 3x + 2 \le {x}^{2} + 2x + 7$

$⇒ 5x \ge - 5$

$⇒ x \ge - 1 . . . . . \left(2\right)$

So from equation \left(1\right) \text{ and } \left(2\right)$we get domain$x \in [ - 1, \infty )