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

1 Answer
Jan 20, 2016

Answer:

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)#