Difference between revisions of "008A Sample Final A, Question 4"

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(Created page with "'''Question:''' Solve. Provide your solution in interval notation. <math>(x-4)(2x+1)(x-1)<0</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundat...")
 
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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! Foundations
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! Foundations: &nbsp;
 
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|1) What are the zeros of the left hand side?
 
|1) What are the zeros of the left hand side?
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! Step 1: &nbsp;
 
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|The zeros of the left hand side are <math>-\frac{1}{2}</math>, 1, and 4
 
|The zeros of the left hand side are <math>-\frac{1}{2}</math>, 1, and 4
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|The zeros split the real number line into 4 intervals: <math>(-\infty, -\frac{1}{2}), (-\frac{1}{2}, 1), (1, 4),</math> and <math>(4, \infty)</math>.
 
|The zeros split the real number line into 4 intervals: <math>(-\infty, -\frac{1}{2}), (-\frac{1}{2}, 1), (1, 4),</math> and <math>(4, \infty)</math>.
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|-
 
|x = 5: (5 - 4)(2(5) + 1)(5 - 1) = (1)(11)(4) = 44 > 0
 
|x = 5: (5 - 4)(2(5) + 1)(5 - 1) = (1)(11)(4) = 44 > 0
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|<table border="1" cellspacing="0" cellpadding="6" align = "center">
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  <tr>
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    <td align = "center"><math> x:</math></td>
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    <td align = "center"><math> x=-1 </math></td>
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    <td align = "center"><math> x= 0 </math></td>
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    <td align = "center"><math> x = 2 </math></td>
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    <td align = "center"><math> x=5 </math></td>
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  </tr>
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  <tr>
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    <td align = "center"><math> f(x):</math></td>
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    <td align = "center"><math> (-) </math></td>
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    <td align = "center"><math> (+) </math></td>
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    <td align = "center"><math> (-) </math></td>
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    <td align = "center"><math> (+) </math></td>
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  </tr>
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</table>
 
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! Step 3:
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! Step 3: &nbsp;
 
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|-
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|We take the intervals for which our test point led to a desired result, (<math>-\infty, -\frac{1}{2}</math>), and (1, 4).
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|We take the intervals for which our test point led the function being negative, (<math>-\infty, -\frac{1}{2}</math>), and (1, 4).
 
|-
 
|-
 
|Since we we are solving a strict inequality we do not need to change the parenthesis to square brackets, and the final answer is <math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>
 
|Since we we are solving a strict inequality we do not need to change the parenthesis to square brackets, and the final answer is <math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>
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! Final Answer: &nbsp;
 
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|<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>
 
|<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>

Latest revision as of 20:35, 27 May 2015

Question: Solve. Provide your solution in interval notation.

Foundations:  
1) What are the zeros of the left hand side?
2) Can the function be both positive and negative between consecutive zeros?
Answer:
1) The zeros are , 1, and 4.
2) No. If the function is positive between 1 and 4 it must be positive for any value of x between 1 and 4.

Solution:

Step 1:  
The zeros of the left hand side are 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}{2}} , 1, and 4
Step 2:  
The zeros split the real number line into 4 intervals: 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, -\frac{1}{2}), (-\frac{1}{2}, 1), (1, 4),} and 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 (4, \infty)} .
We now pick one number from each interval: -1, 0, 2, and 5. We will use these numbers to determine if the left hand side function is positive or negative in each interval.
x = -1: (-1 -4)(2(-1) + 1)(-1 - 1) = (-5)(-1)(-2) = -10 < 0
x = 0: (-4)(1)(-1) = 4 > 0
x = 2: (2-4)(2(2) + 1)(2 - 1) = (-2)(5)(1) = -10 < 0
x = 5: (5 - 4)(2(5) + 1)(5 - 1) = (1)(11)(4) = 44 > 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= 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 = 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=5 }
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):} 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 (+) } 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 (+) }
Step 3:  
We take the intervals for which our test point led the function being negative, (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, -\frac{1}{2}} ), and (1, 4).
Since we we are solving a strict inequality we do not need to change the parenthesis to square brackets, and the final answer 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, -\frac{1}{2}) \cup (1, 4)}
Final Answer:  
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, -\frac{1}{2}) \cup (1, 4)}

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