Question 252676
<font size = 8 color = "red">Note: The other tutor mistakenly did f/g, not g/f.
Solution by Edwin:
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{{{"(g/f)(x)" = (x-2)/(3x-6)}}},{{{x<>2}}}
<pre><font size = 4 color = "indigo"><b>
This function g/f is not defined when the denominator 
equals 0. The denominator would equal 0 when {{{3x-6 = 0}}} 
which simplifies to when {{{x=2}}}.  Therefore
x cannot equal 2, and 2 cannot be part of the domain
of g/f. However when {{{x<>2}}}, the right side 

{{{(x-2)/(3x-6)}}} can be simplified to

{{{(x-2)/(3(x-2))=(cross((x-2)))/(3cross((x-2)))= 1/3}}}

However that cancellation cannot be made when {{{x=2}}}!,
since the function is undefined there, but it can be done
for every other value of x.

So therefore we can express g/f in its simplest form 
this way:

{{{"(g/f)(x)"=1/3}}}, {{{x<>2}}}

Its graph is the green graph below.  It is the horizontal
green line with a hole in it where the point (2,{{{1/3}}})
is missing and that point is NOT part of the graph of g/f:

{{{drawing(400,400,-4,4,-4,4, graph(400,400,-4,4,-4,4),
green(line(-5,1/3,1.92,1/3)), green(line(2.08,1/3,5,1/3)),
 locate(1.92,1/3+.185,o) )}}} 

So the domain of g/f is the x-axis without a value at 2.

In set-builder notation the domain of g/f would be
written {{{"{x|"}}}{{{x<>2}}}{{{"}"}}}.  

The graph of its domain on a number line would look just 
like the x-axis above only, shaded everywhere except at 2,
and with an open circle at 2, like this:

<========================o========>
-4  -3  -2  -1   0   1   2   3   4  

In interval notation this domain is written

{{{"("}}}{{{-infinity}}}{{{",2)"}}}{{{U}}}{{{"(2,"}}}{{{infinity}}}{{{")"}}}

Edwin</pre>