How do you find the domain and range of #sqrt(8-x)#?

2 Answers
Mar 20, 2017

Domain: #[8 , -oo)# , Range: #[0, oo) #

Explanation:

#y= sqrt(8-x) # , Domain : # 8-x >= 0 :. 8 >= x or x <= 8 :.# Domain: #[8 , -oo)#

Range : #y >= 0 :.# Range : #[0, oo) # graph{(8-x)^0.5 [-10, 10, -5, 5]} [Ans]

Mar 20, 2017

Domain: #x <= 8color(white)("xxx")#or #(-oo,8]#
Range: #[0,+oo)#
#color(white)("XXXXXXXXX")#assuming we are restricted to #RR#, the set of real numbers.

Explanation:

#sqrt(8-x)# is defined (in #RR#) for all values of #x# for which #(8-x) >=0#;
that is for #x <= 8#. [This gives us our "Domain"].

#sqrt("anything")# is defined as the primary root i.e. a value #>=0#
At #x=8#, #color(white)("XXXXXXX")sqrt(8-x)=0#
and as #xrarr-oo#, #color(white)("XX")sqrt(8-x)rarr+oo#
[This gives us our "Range"].