Difference between revisions of "005 Sample Final A, Question 2"
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|1) The domain is <math>(0, \infty)</math>. The domain of <math>\frac{1}{x}</math> is <math>[0, \infty)</math>, but we have to remove zero from the domain since we cannot divide by 0. | |1) The domain is <math>(0, \infty)</math>. The domain of <math>\frac{1}{x}</math> is <math>[0, \infty)</math>, but we have to remove zero from the domain since we cannot divide by 0. | ||
|- | |- | ||
| − | |2) <math>x^2 - x -2 = (x - 2)(x - 1)</math> | + | |2) <math>x^2 - x -2 = (x - 2)(x + 1)</math> |
| + | |} | ||
| + | |||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Step 1: | ||
| + | |- | ||
| + | |We start by factoring <math>x^2 - x - 2</math> into <math>(x - 2)(x + 1)</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! 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> | ||
| + | |- | ||
| + | |<table border="1" cellspacing="0" cellpadding="6" align = "center"> | ||
| + | <tr> | ||
| + | <td align = "center"><math> x:</math></td> | ||
| + | <td align = "center"><math> x<-1 </math></td> | ||
| + | <td align = "center"><math> x=-1 </math></td> | ||
| + | <td align = "center"><math> -1<x<2 </math></td> | ||
| + | <td align = "center"><math> x=2 </math></td> | ||
| + | <td align = "center"><math>2<x</math></td> | ||
| + | </tr> | ||
| + | <tr> | ||
| + | <td align = "center"><math> Sign: </math></td> | ||
| + | <td align = "center"><math> (+) </math></td> | ||
| + | <td align = "center"><math> 0 </math></td> | ||
| + | <td align = "center"><math> (-) </math></td> | ||
| + | <td align = "center"><math> 0 </math></td> | ||
| + | <td align = "center"><math> (+)</math></td> | ||
| + | </tr> | ||
| + | </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: |