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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!!!

enter image description here

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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 $(-\infty, -2] \cup (1, \infty)$.

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One thing that should be noted is that strict inequality is not necessary. The square root of 0 exists, so maximal domain should include the point x=-2

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  • $\begingroup$ Well, you got a point but that also would be wrong anyways, take a look at the graph, my answer is incorrect $\endgroup$ Oct 17 at 9:59

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