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

1 Answer
Feb 1, 2018

Answer:

#D_f = {x <= 8 }# and

#R_f = { 0<=y<∞} # where # y = (8-x)^(1/2)#

Explanation:

Since the given function is a square root function so the only NON-NEGATIVE values are allowed in the Domain and the Range will also be having NON-NEGATIVE Real numbers only.

For Domain, #D_f# :

#8-x>=0#
#rArrx-8<=0#
#rArrx <= 8#

and we can see that this function will acquire every positive value and Zero also (when #x=8#).

Thus the Range of function, #R_f = { 0<=y<∞}#