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 1: | ||
+ | |- | ||
+ | |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<5 </math></td> | ||
+ | <td align = "center"><math> x=5 </math></td> | ||
+ | <td align = "center"><math>x>5</math></td> | ||
+ | </tr> | ||
+ | <tr> | ||
+ | <td align = "center"><math> f'(x):</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> | ||
|} | |} |
Revision as of 10:42, 6 May 2015
Question Find the domain of the following function. Your answer should be in interval notation
Foundations: |
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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 1: | ||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|
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 | ||||||||||||