How do you find the domain and range of $\frac{\sqrt{x + 4}}{x + 2}$ $sqrt(x+4)/(x+2)$ ?

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Answer 1 · 2018-07-01T11:31:50

The domain is $x \in [- 4, 2) \cup (2, + \infty)$ $x in [-4,2) uu (2,+oo)$ . The range is $y in RR$

For the domain, there are $2$ conditions :

The denominator must be $!=0$

$=>$ , $x+2!=0$

$=>$ , $x !=-2$

And

What's under the $sqrt$ sign must be $>=0$

$=>$ , $x+4>=0$

$=>$ , $x>=-4$

Therefore,

The domain is $x in [-4,-2) uu (-2,+oo)$

For the range, let $y=sqrt(x+4)/(x+2)$

When $x=-4$ , $=>$ , $y=0$

And

$lim_(x->-2^-)y=-oo$

$lim_(x->-2^+)y=+oo$

$lim_(x->+oo)y=0^+$

The range is $y in RR$

graph{sqrt(x+4)/(x+2) [-11.44, 24.6, -8.1, 9.92]}

How do you find the domain and range of sqrt(x+4)/(x+2)√x+4x+2 sqrt(x+4)/(x+2)?