How do you find the domain & range for #f(x)=sqrt(sinx-1)#?

1 Answer
Oct 22, 2015

#domf={x|x=pi/2+kpi, kin2ZZ}#
#ranf={0}#

Explanation:

Domain
The number inside the radical must be greater than or equal to #0#.

#[1]color(white)(XX)sinx-1>=0#

#[2]color(white)(XX)sinx>=1#

However, the possible values of #sin# are only #[-1,1]#. Therefore, for the inequality to be true, #sinx# must be equal to #1#. For #sinx# to be equal to 1, #x# must be #pi/2# or an angle that is coterminal with #pi/2#.

#color(red)(domf={x|x=pi/2+kpi, kin2ZZ}#

Range
Since the only value #sinx# can have is #1#, the only value in the range is #0#:

#[1]color(white)(XX)y=sqrt(sinx-1)#

#[2]color(white)(XX)y=sqrt(1-1)#

#[3]color(white)(XX)y=sqrt(0)#

#[4]color(white)(XX)y=0#

#color(blue)(ranf={0})#