Question

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

Answer 1

Domain: `x in (-8/17,+oo)`
Range: `y in (0,+oo)`

Explanation:

`y=1/sqrt(h(x))`

• Domain

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)`

• Range

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)`