8A F11 Q4 - Revision history

Matt Lee at 22:23, 6 April 2015

2015-04-06T22:23:06Z

Matt Lee at 22:21, 6 April 2015

2015-04-06T22:21:23Z

Matt Lee at 08:17, 26 March 2015

2015-03-26T08:17:31Z

Matt Lee: Created page with "'''Question:''' Solve. Provide your solution in interval notation. $(x-4)(2x+1)(x-1)<0$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundat..."

2015-03-22T05:50:48Z

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..."

New page

'''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;"
! 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 <math>-\frac{1}{2}</math>, 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:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|The zeros of the left hand side are <math>-\frac{1}{2}</math>, 1, and 4
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|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>.
|-
|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
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|We take the intervals for which our test point led to a desired result, (<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>
|}

← Older revision		Revision as of 22:23, 6 April 2015
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−	[[8AF11Final\|<u>'''Return to Exam</u>''']]	+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

← Older revision		Revision as of 22:21, 6 April 2015
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	\|<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>		\|<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math>
	\|}		\|}
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		+	[[8AF11Final\|<u>'''Return to Exam</u>''']]

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 |-
 |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>
 |}

8A F11 Q4 - Revision history

Matt Lee at 22:23, 6 April 2015

Matt Lee at 22:21, 6 April 2015

Matt Lee at 08:17, 26 March 2015

Matt Lee: Created page with "'''Question:''' Solve. Provide your solution in interval notation. (x-4)(2x+1)(x-1)<0 {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundat..."

Matt Lee: Created page with "'''Question:''' Solve. Provide your solution in interval notation. $(x-4)(2x+1)(x-1)<0$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundat..."