Difference between revisions of "005 Sample Final A, Question 2"
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</tr> | </tr> | ||
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| − | <td align = "center"><math> Sign: </math></td> | + | <td align = "center"><math> \text{Sign: }</math></td> |
<td align = "center"><math> (+) </math></td> | <td align = "center"><math> (+) </math></td> | ||
<td align = "center"><math> 0 </math></td> | <td align = "center"><math> 0 </math></td> | ||
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| The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | ||
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| − | [[005 Sample Final A|'''<u>Return to Sample | + | [[005 Sample Final A|'''<u>Return to Sample Exam</u>''']] |
Latest revision as of 09:47, 2 June 2015
Question Find the domain of the following function. Your answer should be in interval notation
| Foundations: |
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| 1) What is the domain of ? |
| 2) How can we factor ? |
| Answer: |
| 1) The domain is . The domain of is , but we have to remove zero from the domain since we cannot divide by 0. |
| 2) |
| Step 1: |
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| We start by factoring into |
| Step 2: | ||||||||||||
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| Since we cannot divide by zero, and we cannot take the square root of a negative number, we use a sign chart to determine when | ||||||||||||
| Step 3: |
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| Now we just write, in interval notation, the intervals over which the denominator is positive. |
| The domain of the function is: |
| Final Answer: |
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| The domain of the function is: |