Difference between revisions of "005 Sample Final A, Question 2"
Jump to navigation
Jump to search
| (2 intermediate revisions by the same user not shown) | |||
| Line 23: | Line 23: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step | + | ! Step 2: |
|- | |- | ||
|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 <math>(x - 2)(x + 1) > 0</math> | |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 <math>(x - 2)(x + 1) > 0</math> | ||
| Line 34: | Line 34: | ||
<td align = "center"><math> -1<x<2 </math></td> | <td align = "center"><math> -1<x<2 </math></td> | ||
<td align = "center"><math> x=2 </math></td> | <td align = "center"><math> x=2 </math></td> | ||
| − | <td align = "center"><math>x | + | <td align = "center"><math>2<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> 0 </math></td> | <td align = "center"><math> 0 </math></td> | ||
| Line 45: | Line 45: | ||
</tr> | </tr> | ||
</table> | </table> | ||
| + | |} | ||
| + | |||
| + | {| 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 positive. | ||
| + | |- | ||
| + | | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Final Answer: | ||
| + | |- | ||
| + | | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | ||
|} | |} | ||
Latest revision as of 21:31, 21 May 2015
Question Find the domain of the following function. Your answer should be in interval notation
| Foundations: |
|---|
| 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: |
|---|
| We start by factoring into |
| Step 2: | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 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: |
|---|
| Now we just write, in interval notation, the intervals over which the denominator is positive. |
| The domain of the function is: |
| Final Answer: |
|---|
| The domain of the function is: |