 |
|---|
SELECT PAGE NO.
No Books/Pages Are Available |
SELECT PROBLEM NO. FOR THE PAGE |
| PAGE: 222 | SET: Exercises | PROBLEM: 29 |
(a)
Thu function is 
.
Differentiate 
on each side with respect to 
.
Find the critical points.
Since 
is a polynomial it is continuous at all the point.
Thus, the critical points exist when 
.
Equate 
to zero.
and 
.
The critical points are 
and 
.
The test intervals are 
, 
and 
.
| Interval | Test Value | Sign of |
Conclusion |
| |
 |
|
Increasing |
| |
 |
|
Decreasing |
| |
 |
|
Increasing |
The function is increasing on the intervals and .
The function is decreasing on the interval .
(b)
Find the local maximum and local minimum.
The function 
has a local maximum at , because 
changes its sign from positive to negative.
Substitute in 
.
Local maximum is 
.
The function has a local minimum at , because changes its sign from negative to positive.
Local minimum is 
.
(c)
.
Differentiate on each side with respect to 
.
Find the inflection points.
Equate 
to zero.
The inflection point is .
Substitute 
in .
The test intervals are 
and 
.
|
Interval |
Test Value | Sign of  |
Concavity |
|
 |
 |
Down |
|
 |
 |
Up |
The graph is concave up on the interval .
The graph is concave down on the interval .
The inflection point is 
.
(d)
Graph :
Graph the function :
(a)
Increasing on the intervals 
and 
.
Decreasing on the interval .
(b)
Local maximum is 
.
Local minimum is 
.
(c)
Concave up in the interval 
.
Concave down in the interval 
.
Inflection point is .
(d)
Graph of the function is
.
|
"I want to tell you that our students did well on the math exam and showed a marked improvement that, in my estimation, reflected the professional development the faculty received from you. THANK YOU!!!" June Barnett |
|
"Your site is amazing! It helped me get through Algebra." Charles |
|
"My daughter uses it to supplement her Algebra 1 school work. She finds it very helpful." Dan Pease |
Simply chose a support option
  
|