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)
