What is the domain and range of $y = \frac{1}{\sqrt{17 x + 8}}$ $y =1/sqrt(17x+8)$ ?

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Answer 1 · 2016-01-20T23:28:41

Domain: $x \in (- \frac{8}{17}, + \infty)$ $x in (-8/17,+oo)$
Range: $y \in (0, + \infty)$ $y in (0,+oo)$

$y = \frac{1}{\sqrt{h (x)}}$ $y=1/sqrt(h(x))$

The existence conditions are:

${(sqrt(h(x))!=0),(h(x)>=0):} => {(h(x)!=0),(h(x)>=0):} => h(x)>0$

$:.17x+8>0=>x> -8/17$

$:.$ Domain: $x in (-8/17,+oo)$

we have to evaluate:

$lim_(x rarr (-8/17)^+) f(x)=1/0^+=+oo$
$lim_(x rarr (+oo)) f(x)=1/(+oo)=0^+$
```
 then #y=0# is a horizontal asymptote for #x rarr +oo#
```

$:.$ Range: $y in (0,+oo)$

What is the domain and range of y =1/sqrt(17x+8)y=1√17x+8y =1/sqrt(17x+8)?