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| PAGE: 243 | SET: Exercises | PROBLEM: 48 |
The function is 
.
Find the horizontal asymptote.
Therefore the horizontal asymptote is at 
.
To find the vertical asymptote, equate denominator of the function to zero.
So 
Here the roots of the function is imaginary, so there is no vertical asymptote.
The function is 
.
Apply derivative on each side with respect to 
.
Find the critical points.
Thus, the critical points exist when 
.
The critical point is 
.
The test intervals are 
and 
.
| Interval | Test Value | Sign of  |
Conclusion |
 |
 |
|
Decreasing |
 |
 |
|
Increasing |
The function is increasing on the interval .
The function is decreasing on the interval .
Find the local maximum and local minimum.
The function 
has a local minimum at 
, because 
changes its sign from negative to positive.
Substitute in 
.
Local minimum is 
.
.
Again apply derivative on each side with respect to .
Find the inflection points.
Equate 
to zero.
The inflection point is at 
and 
.
The test intervals are 
, 
and 
.
|
Interval |
Test Value | Sign of |
Concavity |
 |
 |
 |
Down |
 |
 |
 |
Up |
 |
 |
 |
Down |
The graph is concave up on the interval .
The graph is concave down on the intervals and .
The inflection points are 
and 
.
Graph :
Graph the function :
The horizontal asymptote is at .
The function is increasing on and decreasing on .
The graph is concave up on and concave down on and .
Graph of the function is
.
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