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

Answer 1

Avery Alexander

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

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

The prerequisites for existence 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 must assess:

#lim_(x rarr (-8/17)^+) f(x)=1/0^+=+oo#

#lim_(x rarr (+oo)) f(x)=1/(+oo)=0^+#

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

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Answer 2

Genesis Allen

Domain: ( x \in \left(-\infty, -\frac{8}{17}\right) \cup \left(-\frac{8}{17}, \infty\right) )

Range: ( y \in \left(0, \infty\right) )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.