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 20:31, 21 May 2015

Question Find the domain of the following function. Your answer should be in interval notation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \frac{1}{\sqrt{x^2-x-2}}}

Foundations:
1) What is the domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{x}}} ?
2) How can we factor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - x - 2} ?
Answer:
1) The domain is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty)} . The domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{x}} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0, \infty)} , but we have to remove zero from the domain since we cannot divide by 0.
2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - x -2 = (x - 2)(x + 1)}


Step 1:
We start by factoring Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - x - 2} into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - 2)(x + 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - 2)(x + 1) > 0}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x:} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<-1 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1<x<2 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=2 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2<x}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Sign: } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (+) } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-) } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 } Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (+)}
Step 3:
Now we just write, in interval notation, the intervals over which the denominator is positive.
The domain of the function is: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\infty, -1) \cup (2, \infty)}
Final Answer:
The domain of the function is: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-\infty, -1) \cup (2, \infty)}
Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_2&oldid=827"