Domain of arcsin[(x^2+1)/2x]

Question

According to me, it should go as follows. $$-1\leq\frac{x^2+1}{2x} \leq1$$

Answer 1

we have:

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

Assume $x>0$ we then have:

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

$$-4x \leq x^2-2x+1 \leq 0$$

$$-4x \leq (x-1)^2 \leq 0$$

The only case this is, is when $x=1$. First consider $(x-1)^2 \leq 0$ and then the other inequality.

Assume $x<0$

Multiply both sides of the inequalities by $2x$, switching the signs because $x<0$

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

Subtract $x^2+1$ from both sides.

$$-2x-x^2-1 \geq 0 \geq -x^2+2x-1$$

Factor.

$$-(x+1)^2 \geq 0 \geq -(x-1)^2$$

Let's take a look at one of the inequalities above $-(x+1)^2 \geq 0$.

This sign chart shows the sign of $-(x+1)^2$ as $x$ varies along the real number line.

How did I get this chart and why does it show what it shows.

First I considered when $-(x+1)^2$ is $0$ and represented that on the number line. I made sure to include all real numbers such that $-(x+1)^2$ is $0$. Then, if $x$ not $0$ it's either positive or negative. One side of where $0$ is has to be either positive or negative, there can't be both negative and positive values to one side of $x=1$ because then that would imply that we had to go through another $0$ by rolles theorem. But I already took care of all $0$'s. So all I need to do is plug in a point to the left of $x=1$ and see what sign it is there, and plug in a point to the right of $x=1$ and see what sign it is there.

So when is $x$ non-negative? What about our constraint? What about our other inequality (again draw a sign chart)? .. Take the intersection of all these constraints to get what we are looking for.

If you do this you will see the two inequalities, along with the constraint, only work at the same time when $x=-1$

So in conclusion (after one considers $x=0$) we have:

$$x \in \{-1,1\}$$

Answer 2

Multiplying through by $\;2x^2\ge0\;$ (this is one of the best tricks to avoid problems with negatives when working with inequalities...) :

$$-1\le\frac{x^2+1}{2x}\le1\iff -2x^2\le x(x^2+1)\le2x^2$$

and now solve each inequality, say the left one:

$$-2x^2\le x(x^2+1)\implies x\left[x^2+1+2x\right]\ge0\iff x(x+1)^2\ge0$$

which is true for any $\;x\ge0\;,\,\color{red}{x=-1}$ , and now the other one:

$$x(x^2+1)\le2x^2\implies x\left[x^2+1-2x\right]\le0\iff x(x-1)^2\le0$$

which is true for $\;x\le 0,\,\color{red}{x=1}\;$, and since $\;x=0\;$ doesn't play with us this time, the solution is, perhaps a little surprisingly, merely $\;\{-1,\,1\}$

Note: observe both quantities in red are the only ones common to both solutions sets, which is precisely what we need when solving an "and" system of inequalities.