Let f(x) = sin^{-1} x and g(x) = frac{x^2 - x | vedclass

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(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.First,we find $g(2) = \lim_{x 	o 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \l

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$(-\infty, -2] \cup [-\frac{4}{3}, \infty)$

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(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.First,we find $g(2) = \lim_{x 	o 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \lim_{x 	o 2} \frac{x+1}{2x+3} = \frac{3}{7}$.Thus,we solve $|\frac{x+1}{2x+3}| \leq 1$ for $x 
eq 2$.This implies $-1 \leq \frac{x+1}{2x+3} \leq 1$.Case $1$: $\frac{x+1}{2x+3} \leq 1 \Rightarrow \frac{x+1 - (2x+3)}{2x+3} \leq 0 \Rightarrow \frac{-x-2}{2x+3} \leq 0 \Rightarrow \frac{x+2}{2x+3} \geq 0$.The solution is $x \in (-\infty, -2] \cup (-\frac{3}{2}, \infty)$.Case $2$: $\frac{x+1}{2x+3} \geq -1 \Rightarrow \frac{x+1 + 2x+3}{2x+3} \geq 0 \Rightarrow \frac{3x+4}{2x+3} \geq 0$.The solution is $x \in (-\infty, -\frac{4}{3}] \cup (-\frac{3}{2}, \infty)$.Taking the intersection of both cases and excluding $x=2$ (since $g(2)$ is defined as the limit),we get $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$,then the domain of the function $f \circ g$ is .... .

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