005 Sample Final A, Question 2 - Revision history

MathAdmin at 17:47, 2 June 2015

2015-06-02T17:47:32Z

MathAdmin at 01:12, 1 June 2015

2015-06-01T01:12:29Z

MathAdmin at 01:11, 1 June 2015

2015-06-01T01:11:41Z

MathAdmin: Created page with "'''Question''' Find the domain of the following function. Your answer should be in interval notation $f(x) = \frac{1}{\sqrt{x^2-x-2}}$
{| class="mw-collapsib..."

2015-06-01T01:11:21Z

Created page with "'''Question''' Find the domain of the following function. Your answer should be in interval notation <math> f(x) = \frac{1}{\sqrt{x^2-x-2}}</math> <br> {| class="mw-collapsib..."

New page

'''Question''' Find the domain of the following function. Your answer should be in interval notation <math> f(x) = \frac{1}{\sqrt{x^2-x-2}}</math> <br>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:
|-
|1) What is the domain of <math>\frac{1}{\sqrt{x}}</math>?
|-
|2) How can we factor <math>x^2 - x - 2</math>?
|-
| Answer:
|-
|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>
|}

{| 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>
|}
[[005 Sample Final A|Return to Sample Final]]

@@ Line 60: / Line 60: @@
 | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math>
 |}
-[[005 Sample Final A|'''<u>Return to Sample Final</u>''']]
+[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

@@ Line 37: / Line 37: @@
    </tr>
    <tr>
-     <td align = "center"><math> Sign: </math></td>
+     <td align = "center"><math> \text{Sign: }</math></td>
      <td align = "center"><math> (+) </math></td>
      <td align = "center"><math> 0 </math></td>

005 Sample Final A, Question 2 - Revision history

MathAdmin at 17:47, 2 June 2015

MathAdmin at 01:12, 1 June 2015

MathAdmin at 01:11, 1 June 2015

MathAdmin: Created page with "'''Question''' Find the domain of the following function. Your answer should be in interval notation f(x) = \frac{1}{\sqrt{x^2-x-2}} {| class="mw-collapsib..."

MathAdmin: Created page with "'''Question''' Find the domain of the following function. Your answer should be in interval notation $f(x) = \frac{1}{\sqrt{x^2-x-2}}$
{| class="mw-collapsib..."