Domain of an inverse sine rational expression

Expert intro: This problem uses the $[-1,1]$ condition for $\sin^{-1}$ and then rewrites it as a quadratic inequality.

Detailed walkthrough

Step 1: Use the inverse sine domain rule

For

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

the input to $\sin^{-1}$ must satisfy

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

A cleaner way is to use absolute value:

$\left|\frac{x-1}{2x+3}\right|\le 1,$ which is equivalent to

$|x-1|\le |2x+3|.$

Step 2: Square both sides

Since both sides are nonnegative, squaring preserves the inequality:

$(x-1)^2\le (2x+3)^2.$

Expand both sides:

$x^2-2x+1\le 4x^2+12x+9.$

Bring everything to one side:

$0\le 3x^2+14x+8.$

Factor:

$3x^2+14x+8=(3x+2)(x+4).$

So we need

$(3x+2)(x+4)\ge 0.$

Step 3: Determine the intervals

The critical points are

$x=-4 \quad\text{and}\quad x=-\frac23.$

Because the quadratic opens upward, the expression is nonnegative outside the roots:

$x\le -4 \quad\text{or}\quad x\ge -\frac23.$

That means the domain is

$\mathbb{R}-\left(-4,-\frac23\right).$

Step 4: Match the requested form

If the domain is written as

$\mathbb{R}-(\alpha,\beta),$

then

$\alpha=-4,\qquad \beta=-\frac23.$

Therefore,

$12\alpha\beta=12\left(-4\right)\left(-\frac23\right)=32.$

Key takeaway

For rational inputs inside $\sin^{-1}$, the fastest path is often to convert the condition to an absolute value inequality and then to a quadratic inequality. That avoids sign confusion from the denominator.

💡 Pitfall guide

A common mistake is to forget that the denominator $2x+3$ cannot be zero. In this problem, $x=-\tfrac32$ is automatically excluded because it lies inside the removed interval, but in other problems a denominator zero can be an endpoint or a separate excluded point. Another error is to solve $-1\le u\le 1$ as two independent inequalities and combine them incorrectly; writing $|u|\le 1$ is usually safer and cleaner.

🔄 Real-world variant

If the expression were

$\sin^{-1}\left(\frac{x-1}{2x-3}\right),$

the same method would work, but the critical points would change because the denominator zero moves to $x=\tfrac32$. You would solve

$|x-1|\le |2x-3|,$

square both sides, and then factor the resulting quadratic. The final domain would again be a union of two rays, but with different boundary values and a different excluded interval.

🔍 Related terms

inverse sine domain, rational inequality, absolute value inequality