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.
 
|-
 
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|2) <math>x^2 - x -2 = (x - 2)(x - 1)</math>
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|2) <math>x^2 - x -2 = (x - 2)(x + 1)</math>
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|}
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 +
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 1:
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|-
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|We start by factoring <math>x^2 - x - 2</math> into <math>(x - 2)(x + 1)</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 2:
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|-
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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 <math>(x - 2)(x + 1) > 0</math>
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|-
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|<table border="1" cellspacing="0" cellpadding="6" align = "center">
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  <tr>
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    <td align = "center"><math> x:</math></td>
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    <td align = "center"><math> x<-1 </math></td>
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    <td align = "center"><math> x=-1 </math></td>
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    <td align = "center"><math> -1<x<2 </math></td>
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    <td align = "center"><math> x=2 </math></td>
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    <td align = "center"><math>2<x</math></td>
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  </tr>
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  <tr>
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    <td align = "center"><math> Sign: </math></td>
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    <td align = "center"><math> (+) </math></td>
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    <td align = "center"><math> 0 </math></td>
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    <td align = "center"><math> (-) </math></td>
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    <td align = "center"><math> 0 </math></td>
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    <td align = "center"><math> (+)</math></td>
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  </tr>
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</table>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 3:
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|-
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| Now we just write, in interval notation, the intervals over which the denominator is positive.
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|-
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| The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Final Answer:
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|-
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| 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: