Determining the domain of an inverse sine rational expression

Expert intro: This question is a standard domain test: the fraction inside arcsine must stay between -1 and 1.

Detailed walkthrough

Step 1: Write the arcsine condition

We are given

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

For $\sin^{-1}(u)$ to be defined, the input must satisfy

$-1\le u\le 1.$

So we require

$-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1.$

The denominator $x^2+2x+7=(x+1)^2+6$ is always positive, so we can safely multiply through by it without changing the inequality direction.

Step 2: Convert to an absolute-value inequality

The double inequality is equivalent to

$\left|\frac{x^2-3x+2}{x^2+2x+7}\right|\le 1.$

Since the denominator is positive, this becomes

$|x^2-3x+2|\le x^2+2x+7.$

Now split into two inequalities:

First inequality

$x^2-3x+2\le x^2+2x+7$

which simplifies to

$-5x\le 5 \Rightarrow x\ge -1.$

Second inequality

$-(x^2-3x+2)\le x^2+2x+7$

or

$-x^2+3x-2\le x^2+2x+7.$

Rearrange:

$0\le 2x^2-x+9.$

This is always true because its discriminant is

$(-1)^2-4(2)(9)=1-72<0$

and the leading coefficient is positive.

So the only restriction is

$x\ge -1.$

Hence the domain is

$[-1,\infty).$

Why this is the clean method

The denominator is always positive, which is a strong hint that an absolute-value approach will be efficient. Instead of testing many cases, you only need to convert the arcsine condition into two simple inequalities. The first one gives the actual restriction, while the second is automatically satisfied for all real $x$.

This is a good example of how algebraic simplification can turn a domain problem into a one-line interval result.

💡 Pitfall guide

A common mistake is to set only the numerator between $-1$ and $1$. That ignores the denominator, which changes the size of the fraction and therefore the allowed $x$ values. Another mistake is to assume the denominator might be zero somewhere, but here it never is, since $x^2+2x+7=(x+1)^2+6>0$ for all real $x$. Do not overcomplicate the problem by factoring first; the direct inequality method is much faster and safer.

🔄 Real-world variant

If the function were $\sin^{-1}\!\left(\frac{x^2-3x+2}{x^2-2x+1}\right)$, the domain would be much more delicate because the denominator becomes $(x-1)^2$, which is zero at $x=1$. You would need both the arcsine bound and the exclusion $x\ne 1$. If the numerator were changed to $x^2-4x+4$, the ratio could simplify and the domain could change significantly. These variants show why checking the denominator structure first is useful.

🔍 Related terms

arcsine domain, rational inequality, quadratic denominator positivity