(a)
\The functions are
and
.
Find the domain of
.
Since it is a rational function, the denominator should not be the zero.
\
Thus, the domain of the function is all real numbers except 0 and 2.
\\
\
The function is
.
Since it is a linear function, the domain is all real numbers.
\(b)
\The function is
.
Simplify the function.
\
Since the above function is a linear function, there is no vertical asymptote.
\(c)
\The functions are
and
.
Make the table :
\| \
x \ | \
\
| \
![]() | \
| \
| \
\
| \
![]() | \
| \
0 \ | \
\
| \
0 | \
| \
1 \ | \
\
| \
1 | \
| \
1.5 \ | \
\
| \
1.5 | \
| 2 | \ \
| \
2 | \
| 2.5 | \ \
| \
2.5 | \
| 3 | \ \
| \
3 | \
(d)
\Graph :
\Draw a coordinate plane.
\Graph the functions
and
in the same viewing window. \ \
Graph of the functions
and
is : \ \
=x%5E2(x-2)_by_x%5E2-2x.gif\")
(e)
\Observe the above graph.
\On a graphing utility, the graphs of f and g will look like the same.
\Since there is a finite number of pixels that can be displayed.
\There is no discontinuity of f at
and
.
(a)
\The domain of the function
is all real numbers except 0 and 2.
The domain of
is all real numbers.
(b)
\None.
\(c)
\Completed table is :
\| \
x \ | \
\
| \
![]() | \
| \
| \
\
| \
![]() | \
| \
0 \ | \
\
| \
0 | \
| \
1 \ | \
\
| \
1 | \
| \
1.5 \ | \
\
| \
1.5 | \
| 2 | \ \
| \
2 | \
| 2.5 | \ \
| \
2.5 | \
| 3 | \ \
| \
3 | \
(d)
\Graph :
\=x%5E2(x-2)_by_x%5E2-2x.gif\")
(e)
\On a graphing utility, the graphs of f and g will look like the same.
\Since there is a finite number of pixels that can be displayed.
\There is no discontinuity of f at
and
.