Question

1. what is the range of $f(x) = \frac{1}{x + 2}$?

$(-\infty, -2) \cup (-2, \infty)$
$(-\infty, 0) \cup (0, \infty)$
$(-\infty, 0) \cap (0, \infty)$
$(-\infty, 2) \cup (2, \infty)$

Explanation:

Step1: Analyze the function

The function is $f(x)=\frac{1}{x+2}$. The denominator $x+2$ can be any real number except $0$ (since division by zero is undefined), so $x+2
eq 0 \implies x
eq -2$ (this is the domain, not range).

Step2: Determine possible output values

Let $y = \frac{1}{x+2}$. Solving for $x$: $x+2 = \frac{1}{y}$, so $x = \frac{1}{y} - 2$. For $x$ to be real, $y$ cannot be $0$ (since $\frac{1}{y}$ is undefined when $y=0$).

Step3: Confirm range

All real numbers except $0$ are possible values of $y$. Thus the range is $(-\infty,0) \cup (0,\infty)$.

Answer:

B. $(-\infty,0) \cup (0,\infty)$