Difference between revisions of "005 Sample Final A, Question 5"
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<tr> | <tr> | ||
<td align = "center"><math> x:</math></td> | <td align = "center"><math> x:</math></td> | ||
| − | <td align = "center"><math> x<-1 </math></td> | + | <td align = "center"><math> x<-2 </math></td> |
| + | <td align = "center"><math> x=-2 </math></td> | ||
| + | <td align = "center"><math> -2<x<-1 </math></td> | ||
<td align = "center"><math> x=-1 </math></td> | <td align = "center"><math> x=-1 </math></td> | ||
| − | <td align = "center"><math> -1<x | + | <td align = "center"><math>-1<x</math></td> |
| − | |||
| − | |||
</tr> | </tr> | ||
<tr> | <tr> | ||
<td align = "center"><math> Sign: </math></td> | <td align = "center"><math> Sign: </math></td> | ||
<td align = "center"><math> (+) </math></td> | <td align = "center"><math> (+) </math></td> | ||
| − | <td align = "center"><math> | + | <td align = "center"><math> VA </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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|} | |} | ||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Step 3: | ||
| + | |- | ||
| + | | Now we just write, in interval notation, the intervals over which the denominator is nonnegative. | ||
| + | |- | ||
| + | | The domain of the function is: <math>(-\infty, -2) \cup [-1, \infty)</math> | ||
| + | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
Latest revision as of 21:33, 21 May 2015
Question Solve the following inequality. Your answer should be in interval notation.
| Step 1: |
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| We start by subtracting 2 from each side to get |
| Step 2: | ||||||||||||
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| Step 3: |
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| Now we just write, in interval notation, the intervals over which the denominator is nonnegative. |
| The domain of the function is: |
| Final Answer: |
|---|