A question about the maximal domain of a function So I have got the following equation:
$$h(x) = \sqrt{\frac{1}{x+1}+1}$$
I need to find the maximal domain of the function.
I have tried doing it algebraically:
As this is a square root, $\frac{1}{x+1}+1$, must be greater than $0$ and there is an asymptote at  $x = -1$.
$\frac{1}{x+1}+1 > 0$
$\frac{1}{x+1} > -1$
$ 1 >-x-1$
$1+x>-1$
$\to x>-2$, provided that $x \not= -1$
But this is incorrect $:($
The answers show a different maximal domain. Moreover, I do not know how they got their answer. I need help. Thanks!!!


In what follows, I will assume that $h$ is a real-valued function of a real variable.
The expression
$$\frac{1}{x + 1}$$
is undefined when $x = -1$.  Thus, we require that $x > -1$ or $x < -1$.
Since we cannot take the square root of a negative number, we require that the radicand (the term inside the square root) be nonnegative.  Hence,
\begin{align*}
\frac{1}{x + 1} + 1 & \geq 0\\
\frac{1}{x + 1} & \geq -1
\end{align*}
At this point, you made a mistake.  You have to consider two cases, depending on the sign of $x + 1$.
Case 1: $x > -1$.
If $x > -1$, then $x + 1 > 0$, so the inequality is preserved if we multiply both sides of the inequality
$$\frac{1}{x + 1} \geq -1$$
by $x + 1$, which yields
\begin{align*}
1 & \geq -x - 1\\
x + 1 & \geq -1\\
x & \geq -2
\end{align*}
which is automatically satisfied if $x > -1$.
Case 2: If $x < -1$, then $x + 1 < 0$, so the direction of the inequality is reversed if we multiply both sides of the inequality
$$\frac{1}{x + 1} \geq -1$$
by $x + 1$.  Hence,
\begin{align*}
1 & \leq -x - 1\\
x & \leq -2
\end{align*}
Therefore, the domain of the function is $(-\infty, -2] \cup (1, \infty)$.
