Difference between revisions of "005 Sample Final A, Question 2"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
+
! 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>
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     <td align = "center"><math> x<-1 </math></td>
 
     <td align = "center"><math> 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<5 </math></td>
+
     <td align = "center"><math> -1<x<2 </math></td>
     <td align = "center"><math> x=5 </math></td>
+
     <td align = "center"><math> x=2 </math></td>
     <td align = "center"><math>x>5</math></td>
+
     <td align = "center"><math>2<x</math></td>
 
   </tr>
 
   </tr>
 
   <tr>
 
   <tr>
     <td align = "center"><math> f'(x):</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>
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   </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;"
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! 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: