1 answer

How do you find domain and range for #f(x) = 2 / sqrt(3x-2)#?

Question:

How do you find domain and range for #f(x) = 2 / sqrt(3x-2)#?

Answers

Refer to explanation

Explanation:

In order for the function to have meaning in the set of real numbers
we must have

#3x-2>0=>3x>2=>x>2/3#

hence, the domain is #(2/3,+oo)#

for the range we set #y=f(x)# and we have

#y=2/sqrt(3x-2)=>(sqrt(3x-2))=2/y=>2/y>0=>y>0#

So the range is #(0,+oo)#

.

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