# Question c73f9

May 29, 2017

1. (b)
2. (d)

#### Explanation:

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

Note that $f \left(x\right)$ is defined if $\left(x - 1\right) \left(x - 2\right) \left(3 - x\right) > 0$

We have critical values for $x$ at $x = 1 , x = 2 , \mathmr{and} x = 3$

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:
$\textcolor{w h i t e}{\text{XXX}} x \in \left\{\frac{1}{2} , 1 , \frac{3}{2} , 2 , \frac{5}{2} , 3 , \frac{7}{2}\right\}$

{: (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 $\left(x - 1\right) \left(x - 2\right) \left(3 - x\right) > 0$ are $x < 1$ and $x \in \left(2 , 3\right)$
i.e. answer (b), the Domain is $x \in \left(- \infty , 1\right) \cup \left(2 , 3\right)$

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

Question 2
Given
$\textcolor{w h i t e}{\text{XXX}} g \left(x\right) = 1 - \sqrt{x + 1}$

Clearly $g \left(x\right)$ is defined provided
$\textcolor{w h i t e}{\text{XXX}} \left(x + 1\right) \ge 0$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x \ge - 1$
i.e. the Domain is ${D}_{g} = \left[- 1 , + \infty\right)$

$\sqrt{x + 1}$ 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 ${R}_{g} = \left(- \infty , + 1\right)$