# Domain of Inverse Trig Function

I was asked to find the domain of $$\arcsin[\frac{x^2+1}{2x}]$$ My first step was $$-1\leq\frac{x^2+1}{2x} \leq1$$

What I don't understand is why I cannot cross multiply to get $$-2x\le {1+x}^{2} \le 2x$$ and then solve the inequality? I tried doing this and got the wrong answer.

• You need to separate the problem into two cases: where $x$ is positive, and where $x$ is negative. This is because multiplying by a negative number flips the sign of the inequality. Commented Nov 19, 2019 at 3:26
• Your last inequality hold only for $x>0$. If $x<0$, you should reverse the signs. Commented Nov 19, 2019 at 3:28

In order to find the domain, you need to check every possible real value of $$x$$. This includes values of $$x$$ where $$x\ge 0$$ and where $$x<0$$. When you multiplied all three terms of

$$-1\leq\frac{x^2+1}{2x} \leq1$$

by $$2x$$ to form

$$-2x\le {1+x}^{2} \le 2x$$

you only considered the case where $$x\ge 0$$. Since an inequality flips sign when multiplied by a negative number, multiplying all of the terms of

$$-1\leq\frac{x^2+1}{2x} \leq1$$

by $$2x$$ when $$x<0$$ forms

$$-2x\ge {1+x}^{2} \ge 2x$$

Therefore, you need to break up the analysis into two cases and consider what happens when $$x\ge 0$$ and $$x<0$$.

Case 1: Suppose $$x\ge 0$$. Then

$$-2x\le {1+x}^{2} \le 2x$$

by the second inequality

$${1+x}^{2} \le 2x$$

we see that $$x=1$$ is the only solution.

Case 2: Suppose $$x < 0$$. Then

$$-2x\ge {1+x}^{2} \ge 2x$$

where from the first inequality

$$1+x^2 \le -2x$$

we have that $$x=-1$$ is the only solution.

So, the domain of $$\arcsin\left(\frac{x^2+1}{2x}\right)$$ is $$x\in\{-1,1\}$$.

Let $$x=\tan y$$

$$z=\arcsin\dfrac{1+x^2}{2x}=\arcsin(\csc2y)$$

$$\implies\sin z=\csc2y$$ which is either $$\ge1$$ or $$\le-1$$

But $$-1\le\sin z\le1$$

So, $$\dfrac{1+x^2}{2x}=\pm1$$

Let $$y=\arcsin\dfrac{x^2+1}{2x}$$

$$x^2-2x\sin y+1=0$$

$$x=\sin y\pm i\cos y$$

As $$x$$ is real, $$\cos y$$ must be $$0$$

$$\implies\sin y=\pm1$$