What is the domain and range of #f(x)= sqrt (3+x-2)#?

1 Answer
Mar 15, 2018

Answer:

Domain: #x>= (-1)#. Using interval notation:#[-1,oo)#

Range: #f(x)>= 0#. Using interval notation:#[0,oo)#

Explanation:

Given:

#color(red)(f(x)=sqrt(3+x-2)#

Domain:

The domain of a function refers to a set of input values for which the function is real and defined.

#sqrt(f(x)) = f(x)>=0#

We solve #(3+x-2)>=0#

We will keep the like terms together as a group.

#x-2+3>=0#

#x+1>=0#

Add #(-1)# to both sides to simplify.

#x+1-1>=0-1#

#rArr x>=(-1)#

Hence, the domain is #x>=(-1)#.

Using interval notation we can write the domain as #color(blue)([-1,oo)#

Range:

The range of a radical function is #f(x)>=k, k = 0#

Hence,

Range: #f(x)>=0#

Using interval notation the range can be written as #color(blue)([0, oo)#

Hope it helps.