How do you find the domain and range of $\frac{1}{x + 2}$ $1/(x+2)$ ?

Question

1373 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-16T03:13:23

Domain: $(- \infty, - 2) \cup (- 2, + \infty)$ $(-oo,-2)uu(-2,+oo)$
Range: $(- \infty, 0) \cup (0, + \infty)$ $(-oo,0) uu (0,+oo)$

$f (x) = \frac{1}{x + 2}$ $f(x) = 1/(x+2$

$f (x)$ $f(x)$ is defined $forall x in RR$ except $x=-2$

Hence, the domain of $f(x)$ in interval notation is $(-oo, -2)uu(-2, +oo)$

Consider $lim_(x->-2-) = -oo$

And $lim_(x->-2+) = +oo$

Also, $f(x) != 0$ for any finite $x$

Hence, the range of $f(x)$ is $(-oo,0) uu (0,+oo)$

As can be seen by the graph of $f(x)$ below.

graph{1/(x+2) [-10, 10, -5, 5]}

How do you find the domain and range of 1/(x+2)1x+2 1/(x+2)?