How do you find the domain and range of #f(x)=sqrt (x+7)/(x^2+7)#?

1 Answer
Dec 25, 2017

Answer:

domain#RR={x:x inRR,x>=(-7)}#and range#RR={y:y inRR,y>=0}#

Explanation:

Suppose,y=#f(x)=sqrt(x+7)/(x^2+7)#
If #x<(-7), #the mentioned equation can not be definable.

#sqrt(x+7)/(x^2+7)#=indefinable,when #x<(-7.)##color(brown)[[As.sqrt(-x in RR)=(INDEFINABLE)]]#

Hence,Domain#RR=color(red){{x:x inRR,x>=(-7)}#

For the value of domain set,range of the equation will be greater than or equal #0#

*So,Range#RR=color(blue){{y:y inRR,y>=0}# *