Precalculus 2014
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67
| PAGE: 169 | SET: Exercises | PROBLEM: 67 |
Please look in your text book for this problem Statement
The inequality is 
.
Now, there are two ways this product could be less than or equal to zero.One factor must be negative or zero and one must be positive or zero.
Case 1 :
and 
.
and 
.
The solution set is 
.
Case 2 :
and 
.
and 
.
The solution in interval notation is 
.
There are no values exist in the interval.
The solution set is 
.
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