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

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

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