Difference between revisions of "005 Sample Final A, Question 2"

From Grad Wiki
Jump to navigation Jump to search
(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...")
 
 
(5 intermediate revisions by the same user not shown)
Line 2: Line 2:
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| 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>?
 +
|-  
 +
|2) How can we factor <math>x^2 - x - 2</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:
 +
|-
 +
|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.
 
|-
 
|-
−
|c) False. <math>y = x^2</math> does not have an inverse.
+
|2) <math>x^2 - x -2 = (x - 2)(x + 1)</math>
 +
|}
 +
 
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
! Step 1:
 
|-
 
|-
−
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+
|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:
 
|-
 
|-
−
|e) True.
+
|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>
 
|-
 
|-
−
|f) False.  
+
|<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:
Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_2&oldid=827"