005 Sample Final A, Question 2 - Revision history

Matt Lee at 04:31, 22 May 2015

2015-05-22T04:31:10Z

Matt Lee at 17:51, 6 May 2015

2015-05-06T17:51:51Z

Matt Lee at 17:46, 6 May 2015

2015-05-06T17:46:06Z

Matt Lee at 17:43, 6 May 2015

2015-05-06T17:43:15Z

Matt Lee at 17:42, 6 May 2015

2015-05-06T17:42:13Z

Matt Lee at 16:56, 6 May 2015

2015-05-06T16:56:08Z

Matt Lee: 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..."

2015-04-30T19:17:50Z

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;"
! Final Answers
|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
|-
|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

@@ Line 34: / Line 34: @@
      <td align = "center"><math> -1<x<2 </math></td>
      <td align = "center"><math> x=2 </math></td>
-     <td align = "center"><math>x>2</math></td>
+     <td align = "center"><math>2<x</math></td>
    </tr>
    <tr>

@@ Line 23: / Line 23: @@
 {| 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>
@@ Line 45: / Line 45: @@
    </tr>
 </table>
 |}

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

@@ Line 32: / Line 32: @@
      <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>x>2</math></td>
    </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> 0 </math></td>

@@ Line 12: / Line 12: @@
 |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>
 |}

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Foundations:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+|1) What is the domain of <math>\frac{1}{\sqrt{x}}</math>?
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+| Answer:
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+|2) <math>x^2 - x -2 = (x - 2)(x - 1)</math>
 |}