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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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