Question

The range of the function $f\left(x\right)=\frac{x+2}{\left|x+2\right|}$, x ≠ −2 is
(a) {−1, 1}
(b) {−1, 0, 1}
(c) {1}
(d) (0, ∞)

Answer 1

(a) {−1, 1}

$f\left(x\right)=\frac{x+2}{\left|x+2\right|}$ , x ≠ −2
$$\begin{gathered} \mathrm{Let}y=\frac{x+2}{\left|x+2\right|} \\ \mathrm{For}\left|x+2\right|>0, \\ \mathrm{or}x>-2, \\ y=\frac{x+2}{x+2}=1 \\ \mathrm{For}\left|x+2\right|<0, \\ \mathrm{or}x<-2, \\ y=\frac{x+2}{-(x+2)}=-1 \\ \mathrm{Thus},y=\{-1,1\} \\ \mathrm{or}\mathrm{range}f\left(x\right)=\{-1,1\}. \end{gathered}$$