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;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Foundations | + | ! Foundations: |
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step 1: | + | ! Step 1: |
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step 2: | + | ! Step 2: |
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step 3: | + | ! Step 3: |
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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). | |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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Final Answer: | + | ! Final Answer: |
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|<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math> | |<math>(-\infty, -\frac{1}{2}) \cup (1, 4)</math> | ||
Revision as of 22:50, 25 May 2015
Question: Solve. Provide your solution 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 (x-4)(2x+1)(x-1)<0}
| 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 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. |
| 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: 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 |
| Step 3: |
|---|
| We take the intervals for which our test point led to a desired result, (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)} |