How do you find the domain of #h(x) = (2x^2 + 5 )/ (sqrt(x-2))#?

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Answer 1 · 2018-06-02T16:55:49

Domain: #{x|x>2}# or in interval notation #(2, oo)#

#h(x) = (2x^2 + 5 )/ (sqrt(x-2))#

The domain is restricted by the radical; what is within it cannot be negative and the denominator; it cannot equal zero:

#x-2>=0#

#x>=2#

and

#sqrt(x-2) !=0#

#(sqrt(x-2))^2 !=0^2#

#x-2 !=0#

#x!=2#

put it together:

#x>2#

Domain: #{x|x>2}# or in interval notation #(2, oo)#