Find the domain of the function f(x) = 1 / sqrt(x + 2). Give your answer in interval notation.

Step-by-step solution

1. 1

   List every restriction the formula imposes, not just the obvious one. There are two distinct hazards stacked on top of each other here, and taking only the first is the classic error. The square root demands a nonnegative radicand, and the fraction demands a nonzero denominator. Both must hold at once.

2. 2

   Write the square-root condition. For $\sqrt{x+2}$ to be a real number,

   $x+2\geq 0$

   On its own this would allow $x=-2$.

3. 3

   Write the denominator condition. The denominator is $\sqrt{x+2}$, and a fraction is undefined when its denominator is zero, so

   $\sqrt{x+2}\neq 0\quad\Longleftrightarrow\quad x+2\neq 0$

   This is exactly what rules out $x=-2$.

4. 4

   Combine the two into one strict inequality. Requiring $x+2\geq 0$ and $x+2\neq 0$ simultaneously leaves

   $x+2>0\quad\Longrightarrow\quad x>-2$

5. 5

   Write the answer in interval notation and spot-check the boundary. The set $x>-2$ is

   $(-2,\infty)$

   The parenthesis at $-2$ is essential: at $x=-2$ the formula becomes $1/\sqrt{0}=1/0$, which is undefined, while at $x=-1.99$ it evaluates to $1/\sqrt{0.01}=10$, a perfectly good value. So the domain starts just to the right of $-2$ and runs to infinity.