Science:Math Exam Resources/Courses/MATH110/December 2012/Question 01 (b)/Solution 2

If you've forgotten the rules used in the first solution (or never learned them), you can also facts about the domain of $\ln(x)$ to determine where the graph should go.

The domain of the function $f(x)=\ln(x)$ is $x>0$ which can be seen in the graph below.

The graph of $g(x)=\ln(1-x)$ will have a similar shape, but has a different domain. Let's test some values to see where the domain of this function is.

x-value		y-value
-2	$\ln(1-(-2))=\ln(3)$	$\ln(3)$
-1	$\ln(1-(-2))=\ln(2)$	$\ln(2)$
0	$\ln(1-0)=\ln(1)$	0
1	$\ln(1-1)=\ln(0)$	undefined
2	$\ln(1-2)=\ln(-1)$	undefined

So clearly this function is defined on the domain $x<1$,

suggesting the graph looks something like this: