Question #c73f9

1 Answer
May 29, 2017

Answer:

  1. (b)
  2. (d)

Explanation:

Question 1
Given
#color(white)("XXX")f(x)=(x-1)/(sqrt((x-1)(x-2)(3-x))#

Note that #f(x)# is defined if #(x-1)(x-2)(3-x) >0#

We have critical values for #x# at #x=1, x=2, and x=3#

Consider a collection of values composed of the critical values and some arbitrary values on each side of the critical values.
For demonstration purposes, I have used:
#color(white)("XXX")x in {1/2, 1, 3/2, 2, 5/2, 3, 7/2}#

#{: (ul(x=),"||",ul((x-1)),ul((x-2)),ul((3-x)),"|",ul((x-1)*(x-2)(*3-x))), (1/2,"||",<0,<0,>0,"|",>0), (1,"||",=0,<0,>0,"|",=0), (3/2,"||",>0,<0,>0,"|",<0), (2,"||",>0,=0,>0,"|",=0), (5/2,"||",>0,>0,>0,"|",>0), (3,"||",>0,>0,=0,"|",=0), (7/2,"||",>0,>0,<0,"|",<0) :}#

If you like you could show these on a number line,
but it should be clear that the only values of #x# for which #(x-1)(x-2)(3-x)>0# are #x< 1# and #x in (2,3)#
i.e. answer (b), the Domain is #x in (-oo,1) uu (2,3)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question 2
Given
#color(white)("XXX")g(x)=1-sqrt(x+1)#

Clearly #g(x)# is defined provided
#color(white)("XXX")(x+1) >=0#
#color(white)("XXX")rarr x >= -1#
i.e. the Domain is #D_g=[-1,+oo)#

#sqrt(x+1)# can take on any value greater than or equal to #0#
So the maximum value of #1-sqrt(x+1)# is #1# (when #sqrt(x+1) = 0#)
and can take on any value less than this (when #(sqrt(x+1) > 0#)
i.e. the Range is #R_g=(-oo,+1)#

...that is answer (d)