### Instruction: Let $f(x) = \left\lceil\dfrac{1}{x+2}\right\rceil$ for $x > -2$, and $f(x) = \left\lfloor\dfrac{1}{x+2}\right\rfloor$ for $x < -2$. ($f(x)$ is not defined at $x = -2$.) Which integer is not in the range of $f(x)$?### Response: To find the range of $f(x)$, I need to look at the behavior of $\dfrac{1}{x+2}$ and how the ceiling and floor functions affect it.
For $x > -2$, $\dfrac{1}{x+2}$ is positive and decreasing, approaching zero as $x$ increases.
The ceiling function rounds it up to the next integer, so the smallest possible value of $f(x)$ for $x > -2$ is $f(0) = \left\lceil\dfrac{1}{2}\right\rceil = 1$.
For $x < -2$, $\dfrac{1}{x+2}$ is negative and increasing, approaching zero as $x$ decreases.
The floor function rounds it down to the next integer, so the largest possible value of $f(x)$ for $x < -2$ is $f(-3) = \left\lfloor\dfrac{1}{-1}\right\rfloor = -1$.
Therefore, the range of $f(x)$ is $\{-1, 0, 1\}$ and any integer other than these is not in the range.