Grad Wiki - User contributions [en]

https://gradwiki.math.ucr.edu/api.php?action=feedcontributions&feedformat=atom&user=Matt+Lee Grad Wiki - User contributions [en] 2026-04-23T19:26:02Z User contributions MediaWiki 1.35.0 https://gradwiki.math.ucr.edu/index.php?title=Best_Practices&diff=878 Best Practices 2015-06-04T16:47:34Z

<p>Matt Lee: /* Exam numbering */</p> <hr /> <div>This page covers the simplest way to do different things on the Wiki, or why we use one method over another.<br /> <br /> == Collapsible table ==<br /> Below you will find the code required to make a basic table with helpful descriptions informing you what the code on each line does:<br><br /> <code> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" <br><br /> <br /> The "{|" is required to start a table. The next two pieces of code in quotes make the table collapsible and left justify the text.<br><br /> ! Header of table. This is always visible.<br><br /> |- The code in this line generates a new row. You can modify the text of the row by using html code.<br><br /> |Text for the row generated by the previous row goes here. <br><br /> |}</code> The "|}" closes the table<br /> == Alignment ==<br /> blah<br /> ==Foundations/Solutions ==<br /> For every practice exam you will see a foundations collapsible table, solutions collapsible table, and final answer collapsible table. <br><br /> The Foundations table is for mentioning background material required to understand the solution <br><br /> The solution is a step by step solution to the problem <br><br /> The final answer is just the final answer. The idea behind providing just the final answer is for students who just want to know <br><br /> if they did everything correctly.<br /> <br /> Please stick to this format as it will provide with useful information that we can use to improve the wiki.<br /> <br /> ==Exam numbering==<br /> If the exam was provided by the department it is labeled with a number, eg. Department Sample Midterm 3. If it was a sample midterm given by a professor it is labeled with a letter, eg. Sample Midterm C.</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=Best_Practices&diff=877 Best Practices 2015-06-04T16:47:04Z

<p>Matt Lee: </p> <hr /> <div>This page covers the simplest way to do different things on the Wiki, or why we use one method over another.<br /> <br /> == Collapsible table ==<br /> Below you will find the code required to make a basic table with helpful descriptions informing you what the code on each line does:<br><br /> <code> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" <br><br /> <br /> The "{|" is required to start a table. The next two pieces of code in quotes make the table collapsible and left justify the text.<br><br /> ! Header of table. This is always visible.<br><br /> |- The code in this line generates a new row. You can modify the text of the row by using html code.<br><br /> |Text for the row generated by the previous row goes here. <br><br /> |}</code> The "|}" closes the table<br /> == Alignment ==<br /> blah<br /> ==Foundations/Solutions ==<br /> For every practice exam you will see a foundations collapsible table, solutions collapsible table, and final answer collapsible table. <br><br /> The Foundations table is for mentioning background material required to understand the solution <br><br /> The solution is a step by step solution to the problem <br><br /> The final answer is just the final answer. The idea behind providing just the final answer is for students who just want to know <br><br /> if they did everything correctly.<br /> <br /> Please stick to this format as it will provide with useful information that we can use to improve the wiki.<br /> <br /> ==Exam numbering==<br /> If the exam was provided by the department it is labeled with a number, eg. Sample Midterm 3. If it was a sample midterm given by a professor it is labeled with a letter, eg. Sample Midterm C.</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A&diff=854 005 Sample Final A 2015-05-31T21:20:46Z

<p>Matt Lee: </p> <hr /> <div><span class = "biglink">[[005 Sample Final A, Question 1|&nbsp;Question 1&nbsp;]]</span> Please circle either true or false,<br><br /> &nbsp;&nbsp;&nbsp;&nbsp;a. (True/False)In a geometric sequence, the common ratio is always positive. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp;b. (True/False) A linear system of equations always has a solution. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp;c. (True/False) Every function has an inverse. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp;d. (True/False) Trigonometric equations do not always have unique solutions. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp;e. (True/False) The domain of <math>f(x) = \tan^{-1}(x)</math> is all real numbers. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp;f. (True/False) The function <math> \log_a(x)</math> is defined for all real numbers.<br><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 2|&nbsp;Question 2&nbsp;]]</span> 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><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 3|&nbsp;Question 3&nbsp;]]</span> Find f <math>\circ</math> g and its domain if <math>f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}</math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 4|&nbsp;Question 4&nbsp;]]</span> Find the inverse of the following function <math> f(x) = \frac{3x}{2x-1}</math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 5|&nbsp;Question 5&nbsp;]]</span> Solve the following inequality. Your answer should be in interval notation. <math>\frac{3x+5}{x+2}\ge 2</math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 6|&nbsp;Question 6&nbsp;]]</span> Factor the following polynomial completely, &nbsp;&nbsp;&nbsp;&nbsp;<math>p(x) = x^4 + x^3 + 2x-4 </math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 7|&nbsp;Question 7&nbsp;]]</span> Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 2\log_5(x) = 3\log_5(4)</math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 8|&nbsp;Question 8&nbsp;]]</span> Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 3^{2x} + 3^x -2 = 0 </math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 9|&nbsp;Question 9&nbsp;]]</span> Solve the following system of equations <br><br /> <center><math> \begin{align} 2x + 3y &= & 1\\ -x + y & = & -3\end{align}</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 10|&nbsp;Question 10&nbsp;]]</span> Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 11|&nbsp;Question 11&nbsp;]]</span> Solve the following equation in the interval <math> [0, 2\pi)</math> <br><br /> <center><math> \sin^2(\theta) - \cos^2(\theta)=1+\cos(\theta)</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 12|&nbsp;Question 12&nbsp;]]</span> Given that <math>\sec(\theta) = -2</math> and <math>\tan(\theta) > 0 </math>, find the exact values of the remaining trig functions.<br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 13|&nbsp;Question 13&nbsp;]]</span> Give the exact value of the following if its defined, otherwise, write undefined. <br><br /> <math>(a) \sin^{-1}(2) \qquad \qquad (b) \sin\left(\frac{-32\pi}{3}\right) \qquad \qquad (c)\sec\left(\frac{-17\pi}{6}\right)</math><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 14|&nbsp;Question 14&nbsp;]]</span>Prove the following identity, <br><br /> <center><math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 15|&nbsp;Question 15&nbsp;]]</span> Find an equivalent algebraic expression for the following, <center><math> \cos(\tan^{-1}(x))</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 16|&nbsp;Question 16&nbsp;]]</span> Graph the following, <center><math> -x^2+4y^2-2x-16y+11=0</math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 17|&nbsp;Question 17&nbsp;]]</span> Graph the following function, <center><math>f(x) = \log_2(x+1) + 2</math></center> <br><br /> Make sure to label any asymptotes, and at least two points on the graph.<br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 18|&nbsp;Question 18&nbsp;]]</span> Graph the following function, <center><math>f(x) = \left(\frac{1}{3}\right)^{x+1} + 1</math></center> <br><br /> Make sure to label any asymptotes, and at least two points on the graph.<br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 19|&nbsp;Question 19&nbsp;]]</span> Consider the following function, <center><math>f(x) = -\sin\left(3x+\frac{\pi}{2}\right)+1</math></center><br><br /> <br /> &nbsp;&nbsp;&nbsp;&nbsp; a. What is the amplitude?<br><br /> &nbsp;&nbsp;&nbsp;&nbsp; b. What is the period? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; c. What is the phase shift? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; d. What is the vertical shift? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; e. Graph one cycle of f(x). Make sure to label five key points.<br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 20|&nbsp;Question 20&nbsp;]]</span> Consider the following rational function,<br /> <center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br><br /> <br /> &nbsp;&nbsp;&nbsp;&nbsp; a. What is the domain of f? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; b. What are the x and y-intercepts of f? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes? <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; d. Graph f(x). Make sure to include the information you found above.<br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 21|&nbsp;Question 21&nbsp;]]</span> Find the sum <br><br /> <center><math> 5 + 9 + 13 + \cdots + 49 </math></center><br /> <br /> <span class = "biglink">[[005 Sample Final A, Question 22|&nbsp;Question 22&nbsp;]]</span> Consider the following sequence, <br><br /> <center><math> -3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots </math></center><br><br /> &nbsp;&nbsp;&nbsp;&nbsp; a. Determine a formula for <math>a_n</math>, the n-th term of the sequence. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; b. Find the sum <math> \displaystyle{\sum_{k=1}^\infty a_k}</math></div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&diff=847 022 Sample Final A, Problem 4 2015-05-30T20:16:46Z

<p>Matt Lee: </p> <hr /> <div><br /> <br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp; <br /> |-<br /> |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so <br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.</math><br /> |-<br /> |For this problem we also need to use the product rule.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> and apply the product rule on the right hand side to find that&thinsp; <math style="vertical-align: -18%">x + y = x^3y^3</math> &thinsp;differentiates implicitly to<br /> |-<br /> |<br /> ::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |<math>\frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math><br /> |}<br /> [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&diff=846 022 Sample Final A, Problem 4 2015-05-30T20:16:16Z

<p>Matt Lee: </p> <hr /> <div><br /> <br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp; <br /> |-<br /> |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so <br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.</math><br /> |-<br /> |For this problem we also need to use the product rule.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> and apply the product rule on the right hand side to find that&thinsp; <math style="vertical-align: -18%">x + y = x^3y^3</math> &thinsp;differentiates implicitly to<br /> |-<br /> |<br /> ::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |<math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math><br /> |}<br /> [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&diff=845 022 Sample Final A, Problem 4 2015-05-30T20:06:19Z

<p>Matt Lee: </p> <hr /> <div><br /> <br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp; <br /> |-<br /> |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so <br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math><br /> |}<br /> <br /> &nbsp;'''Solution:'''<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.<br /> |}<br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math> &thinsp;to find<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3(1)^{2}-3(0)^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1,</math><br /> |-<br /> |which is equivalent to <math style="vertical-align:-60%">3-\frac{dy}{dx}=1</math>. This solves to <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&diff=844 022 Sample Final A, Problem 4 2015-05-30T20:05:50Z

<p>Matt Lee: </p> <hr /> <div><br /> <br /> <span style="font-size:135%"><font face=Times Roman><span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp; <br /> |-<br /> |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so <br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math><br /> |}<br /> <br /> &nbsp;'''Solution:'''<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.<br /> |}<br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math> &thinsp;to find<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3(1)^{2}-3(0)^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1,</math><br /> |-<br /> |which is equivalent to <math style="vertical-align:-60%">3-\frac{dy}{dx}=1</math>. This solves to <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&diff=843 022 Sample Final A, Problem 4 2015-05-30T20:05:20Z

<p>Matt Lee: Created page with " <span style="font-size:135%"><font face=Times Roman>4. <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math> {| class="mw..."</p> <hr /> <div><br /> <br /> <span style="font-size:135%"><font face=Times Roman>4. <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp; <br /> |-<br /> |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so <br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math><br /> |}<br /> <br /> &nbsp;'''Solution:'''<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.<br /> |}<br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math> &thinsp;to find<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math>3(1)^{2}-3(0)^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1,</math><br /> |-<br /> |which is equivalent to <math style="vertical-align:-60%">3-\frac{dy}{dx}=1</math>. This solves to <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align:-24%">dy/dx=2.</math><br /> |}<br /> [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=842 022 Sample Final A, Problem 3 2015-05-30T19:58:14Z

<p>Matt Lee: </p> <hr /> <div><span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math><br /> |-<br /> |To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math><br /> |-<br /> |Plugging in -3 for x to both sides we find that <math>6 = -7B</math> &nbsp; and &nbsp; <math>B = -\frac{6}{7}</math>.<br /> |-<br /> |Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math>&nbsp;, so <math>A = \frac{6}{7}</math><br /> |-<br /> |Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 3: &nbsp;<br /> |-<br /> |Now to finish the problem we integrate each fraction to get: <math>\int \frac{6}{x^2 -x -12} dx = \int \frac{6}{7(x - 4)}dx - \int \frac{6}{7(x + 3)}dx </math>&nbsp; to get &nbsp;<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 4: &nbsp;<br /> |-<br /> |Now make sure you remember to add the <math> + C</math> to the integral at the end.<br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math><br /> |}<br /> <br /> <br /> [[022_Sample Final A|'''<u>Return to Sample Final</u>''']]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=841 022 Sample Final A, Problem 3 2015-05-30T19:56:55Z

<p>Matt Lee: </p> <hr /> <div><span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math><br /> |-<br /> |To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math><br /> |-<br /> |Plugging in -3 for x to both sides we find that <math>6 = -7B</math> &nbsp; and &nbsp; <math>B = -\frac{6}{7}</math>.<br /> |-<br /> |Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math>&nbsp;, so <math>A = \frac{6}{7}</math><br /> |-<br /> |Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 3: &nbsp;<br /> |-<br /> |Now to finish the problem we integrate each fraction to get: <math>\int \frac{6}{x^2 -x -12} dx = \int \frac{6}{7(x - 4)}dx - \int \frac{6}{7(x + 3)}dx </math>&nbsp; to get &nbsp;<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 4: &nbsp;<br /> |-<br /> |Now make sure you remember to add the <math> + C</math> to the integral at the end.<br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math><br /> |}<br /> <br /> <br /> [[022_Sample Final A|<span class="biglink"><span style="font-size:80%">&nbsp;Return to Exam&nbsp;</span></span>]]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=840 022 Sample Final A, Problem 3 2015-05-30T19:54:50Z

<p>Matt Lee: </p> <hr /> <div><span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math><br /> |-<br /> |To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math><br /> |-<br /> |Plugging in -3 for x to both sides we find that <math>6 = -7B</math> &nbsp; and &nbsp; <math>B = -\frac{6}{7}</math>.<br /> |-<br /> |Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math>&nbsp;, so <math>A = \frac{6}{7}</math><br /> |-<br /> |Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 3: &nbsp;<br /> |-<br /> |Now to finish the problem we integrate each fraction to get: <math>\int \frac{6}{x^2 -x -12} dx = \int \frac{6}{7(x - 4)}dx - \int \frac{6}{7(x + 3)}dx </math>&nbsp; to get &nbsp;<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 4: &nbsp;<br /> |-<br /> |Now make sure you remember to add the <math> + C</math> to the integral at the end.<br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Final Answer: &nbsp;<br /> |-<br /> |<math>\frac{6}{7}\ln(x - 4) - \frac{6}{7}\ln(x + 3) + C</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=839 022 Sample Final A, Problem 3 2015-05-30T19:43:38Z

<p>Matt Lee: </p> <hr /> <div><span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}<br /> <br /> '''Solution:'''<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 1: &nbsp;<br /> |-<br /> |First, we factor <math>x^2 - x - 12 = (x - 4)(x + 3)</math><br /> |}<br /> <br /> {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Step 2: &nbsp;<br /> |-<br /> |Now we want to find the partial fraction expansion for <math>\frac{6}{(x - 4)(x + 3)}</math>&nbsp;, which will have the form <math>\frac{A}{x - 4} + {B}{x + 3}</math><br /> |-<br /> |To do this we need to solve the equation <math>6 = A( x + 3) + B(x - 4)</math><br /> |-<br /> |Plugging in -3 for x to both sides we find that <math>6 = -7B</math> &nbsp; and &nbsp; <math>B = -\frac{6}{7}</math>.<br /> |-<br /> |Now we can find A by plugging in 4 for x to both sides. This yields <math>6 = 7A</math>&nbsp;, so <math>A = \frac{6}{7}</math><br /> |-<br /> |Finally we have the partial fraction expansion: <math>\frac{6}{x^2 -x - 12} = \frac{6}{7(x - 4)} - \frac{6}{7(x + 3)}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=838 022 Sample Final A, Problem 3 2015-05-30T19:26:07Z

<p>Matt Lee: </p> <hr /> <div><span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_3&diff=837 022 Sample Final A, Problem 3 2015-05-30T19:23:04Z

<p>Matt Lee: Created page with "{| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: &nbsp; |- |1) What does the denominator factor into? What will be the form of the decomposition..."</p> <hr /> <div>{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: &nbsp;<br /> |-<br /> |1) What does the denominator factor into? What will be the form of the decomposition?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> 3) What special integral do we have to use?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since <math>x^2 - x - 12 = (x - 4)(x +3)</math> &nbsp;, and each term has multiplicity one, the decomposition will be of the form: <math>\frac{A}{x - 4} + \frac{B}{x + 3}</math><br /> |-<br /> |2) After writing the equality, <math>\frac{6}{x^2 -x - 12} = \frac{A}{x - 4} + \frac{B}{x + 3}</math>, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.<br /> |-<br /> |3) We have to remember that <math>\int \frac{c}{x - a} dx = c\ln(x - a)</math>&nbsp;, for any numbers c, a.<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=836 022 Sample Final A 2015-05-30T19:09:14Z

<p>Matt Lee: </p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Sample Final A,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Sample Final A,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Sample Final A,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Sample Final A,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Sample Final A,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Sample_Final_A,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Sample_Final_A,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Sample_Final_A,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Sample_Final_A,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demand <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> ::<span class="exam">a) Marginal revenue when x = 7 units.<br /> ::<span class="exam">b) The quantity(x-value) that produces minimum average cost.<br /> ::<span class="exam">c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Sample_Final_A,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">Set up the formula to find the amount of money one would have at the end of 8 years if she invests $2100 in an account paying 6% annual interest, compounded quarterly.<br /> <br /> == [[022_Sample_Final_A,_Problem_11|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 11&nbsp;</span>]] ==<br /> <span class="exam">Find the derivative: <math>g(x) = \frac{ln(x^3 + 7)}{(x^4 + 2x^2)}</math> (note: you do not need to simplify the derivative after finding it)<br /> <br /> == [[022_Sample_Final_A,_Problem_12|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 12&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int x^2e^{3x^3}dx</math><br /> <br /> == [[022_Sample_Final_A,_Problem_13|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 13&nbsp;</span>]] ==<br /> <span class="exam">Use differentials to find <math>dy</math> given <math>y = x^2 - 6x, ~ x = 4, ~dx = -0.5</math><br /> <br /> == [[022_Sample_Final_A,_Problem_14|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 14&nbsp;</span>]] ==<br /> <span class="exam"> Find all vertical and horizontal asymptotes for <math>\qquad y = \frac{x + 4}{(x - 3)(x + 2)}</math><br /> <br /> == [[022_Sample_Final_A,_Problem_15|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 15&nbsp;</span>]] ==<br /> <span class="exam"> Find the following limit: <math>\qquad \lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math></div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=835 022 Sample Final A 2015-05-30T19:07:17Z

<p>Matt Lee: /* &nbsp;Problem 2&nbsp; */</p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Sample Final A,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Sample Final A,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demand <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> ::<span class="exam">a) Marginal revenue when x = 7 units.<br /> ::<span class="exam">b) The quantity(x-value) that produces minimum average cost.<br /> ::<span class="exam">c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">Set up the formula to find the amount of money one would have at the end of 8 years if she invests $2100 in an account paying 6% annual interest, compounded quarterly.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_11|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 11&nbsp;</span>]] ==<br /> <span class="exam">Find the derivative: <math>g(x) = \frac{ln(x^3 + 7)}{(x^4 + 2x^2)}</math> (note: you do not need to simplify the derivative after finding it)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_12|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 12&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int x^2e^{3x^3}dx</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_13|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 13&nbsp;</span>]] ==<br /> <span class="exam">Use differentials to find <math>dy</math> given <math>y = x^2 - 6x, ~ x = 4, ~dx = -0.5</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_14|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 14&nbsp;</span>]] ==<br /> <span class="exam"> Find all vertical and horizontal asymptotes for <math>\qquad y = \frac{x + 4}{(x - 3)(x + 2)}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_15|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 15&nbsp;</span>]] ==<br /> <span class="exam"> Find the following limit: <math>\qquad \lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math></div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=834 022 Sample Final A 2015-05-30T19:06:20Z

<p>Matt Lee: </p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Sample Final A,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demand <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> ::<span class="exam">a) Marginal revenue when x = 7 units.<br /> ::<span class="exam">b) The quantity(x-value) that produces minimum average cost.<br /> ::<span class="exam">c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">Set up the formula to find the amount of money one would have at the end of 8 years if she invests $2100 in an account paying 6% annual interest, compounded quarterly.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_11|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 11&nbsp;</span>]] ==<br /> <span class="exam">Find the derivative: <math>g(x) = \frac{ln(x^3 + 7)}{(x^4 + 2x^2)}</math> (note: you do not need to simplify the derivative after finding it)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_12|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 12&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int x^2e^{3x^3}dx</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_13|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 13&nbsp;</span>]] ==<br /> <span class="exam">Use differentials to find <math>dy</math> given <math>y = x^2 - 6x, ~ x = 4, ~dx = -0.5</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_14|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 14&nbsp;</span>]] ==<br /> <span class="exam"> Find all vertical and horizontal asymptotes for <math>\qquad y = \frac{x + 4}{(x - 3)(x + 2)}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_15|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 15&nbsp;</span>]] ==<br /> <span class="exam"> Find the following limit: <math>\qquad \lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math></div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=833 022 Sample Final A 2015-05-30T19:01:06Z

<p>Matt Lee: </p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demand <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> ::<span class="exam">a) Marginal revenue when x = 7 units.<br /> ::<span class="exam">b) The quantity(x-value) that produces minimum average cost.<br /> ::<span class="exam">c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">Set up the formula to find the amount of money one would have at the end of 8 years if she invests $2100 in an account paying 6% annual interest, compounded quarterly.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_11|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 11&nbsp;</span>]] ==<br /> <span class="exam">Find the derivative: <math>g(x) = \frac{ln(x^3 + 7)}{(x^4 + 2x^2)}</math> (note: you do not need to simplify the derivative after finding it)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_12|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 12&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int x^2e^{3x^3}dx</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_13|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 13&nbsp;</span>]] ==<br /> <span class="exam">Use differentials to find <math>dy</math> given <math>y = x^2 - 6x, ~ x = 4, ~dx = -0.5</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_14|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 14&nbsp;</span>]] ==<br /> <span class="exam"> Find all vertical and horizontal asymptotes for <math>\qquad y = \frac{x + 4}{(x - 3)(x + 2)}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_15|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 15&nbsp;</span>]] ==<br /> <span class="exam"> Find the following limit: <math>\qquad \lim_{x \rightarrow -3}\frac{x^2 + 7x + 12}{x^2 - 2x - 14}</math></div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=832 022 Sample Final A 2015-05-30T18:53:23Z

<p>Matt Lee: </p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demamd <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> ::<span class="exam">a) Marginal revenue when x = 7 units.<br /> ::<span class="exam">b) The quantity(x-value) that produces minimum average cost.<br /> ::<span class="exam">c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">'''Use calculus to set up and solve the word problem:'''<br /> A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4<sup>th</sup> side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=831 022 Sample Final A 2015-05-30T18:51:06Z

<p>Matt Lee: </p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal: <math>\qquad f(x, y) = \frac{2xy}{x-y}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for the her, the pasure must contain 200 square meters of grass. No fencing is required along the river. What dimensions will use the smallest amount of fencing?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative: <math>\int \frac{6}{x^2 - x - 12}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find producer and consumer surpluses for the following situation:<br /> <br /> <math>\text{Supply curve: }\qquad p = 18 + 3x^2 \qquad \text{Demand curve: }\qquad p = 150 - 4x</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the curve, including all relative exterma and points of inflection. <math>y = 3x^4 - 4x^3</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the present value of the income stream <math>Y = 20 + 30x</math> from now until 5 years from now, given an interest rate <math>r = 10%.</math> (note that once you plug in the limits of integration, you are finished- you do not need to simplify our answer beyond that step).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find ther marginial productivity of labor and marginal productivity of capital for the following Cobb-Douglas productio function: (note: you must simplify so your solution does not contain negative exponents)<br /> <br /> <math>f(k, l) = 200k^{0.6}l^{0.4}</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"> Given demamd <math>p = 116 - 3x</math> &nbsp;, and cost <math>C = x^2 + 20x + 64</math>&nbsp;, find:<br /> <br /> <span class="exam">::a) Marginal revenue when x = 7 units.<br /> <span class="exam">::b) The quantity(x-value) that produces minimum average cost.<br /> <span class="exam">::c) Maximum profit (find the x-value and the profit itself)<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">'''Use calculus to set up and solve the word problem:'''<br /> A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4<sup>th</sup> side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A&diff=830 022 Sample Final A 2015-05-30T18:37:01Z

<p>Matt Lee: Created page with "'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class=..."</p> <hr /> <div>'''This is a sample, and is meant to represent the material usually covered in Math 22 for the second exam. An actual test may or may not be similar. Click on the <span class="biglink" style="color:darkblue;">&nbsp;boxed problem numbers&nbsp;</span> to go to a solution.'''<br /> <div class="noautonum">__TOC__</div><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_1|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 1&nbsp;</span></span>]] ==<br /> <span class="exam">Find the derivative of &thinsp;<math style="vertical-align: -60%">y\,=\,\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}.</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_2|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 2&nbsp;</span>]] ==<br /> <span class="exam"> Sketch the graph of <math style="vertical-align: -52%">y = \left(\frac{1}{2}\right)^{x + 1} - 4</math>.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_3|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 3&nbsp;</span>]] ==<br /> <span class="exam"> Find the derivative of <math style="vertical-align: -18%">f(x) \,=\, 2x^3e^{3x+5}</math>.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_4|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 4&nbsp;</span>]] ==<br /> <span class="exam"> '''Set up the equation to solve. You only need to plug in the numbers-not solve for the particular values!'''<br /> <br /> <span class="exam">What is the present value of $3000, paid 8 years from now, in an investment that pays 6%interest,<br /> <br /> ::<span class="exam">(a) compounded quarterly?<br /> ::<span class="exam">(b) compounded continuously?<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_5|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 5&nbsp;</span>]] ==<br /> <span class="exam"> Find the antiderivative of <math>\int \frac{2e^{2x}}{e^2x + 1}\, dx.</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_6|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 6&nbsp;</span>]] ==<br /> <span class="exam">Find the area under the curve of&thinsp; <math style="vertical-align: -13%">y = 6x^2 + 2x</math> between the <math style="vertical-align: -15%">y</math>-axis and <math style="vertical-align: -1%">x = 2</math>.<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_7|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 7&nbsp;</span>]] ==<br /> <span class="exam">Find the antiderivatives:<br /> <br /> ::<span class="exam">(a) <math> \int xe^{3x^2+1}\,dx.</math><br /> <br><br /> ::<span class="exam">(b) <math>\int_2^54x - 5\,dx.</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_8|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 8&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find the quantity that produces maximum profit, given demand function <math style="vertical-align: -15%">p = 70 - 3x</math> and cost function&thinsp; <math style="vertical-align: -8%">C = 120 - 30x + 2x^2.</math><br /> <br /> == [[022_Exam_2_Sample_B,_Problem_9|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 9&nbsp;</span>]] ==<br /> <span class="exam"><br /> Find all relative extrema and points of inflection for the function <math style="vertical-align: -16%">g(x) = x^3 - 3x</math>. Be sure to give coordinate pairs for each point. You do not need to draw the graph. Explain how you know which point is the local minimum and which is the local maximum (i.e., which test did you use?).<br /> <br /> == [[022_Exam_2_Sample_B,_Problem_10|<span class="biglink"><span style="font-size:80%">&nbsp;Problem 10&nbsp;</span>]] ==<br /> <span class="exam">'''Use calculus to set up and solve the word problem:'''<br /> A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4<sup>th</sup> side costs $6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=Pages_For_External_Wiki&diff=829 Pages For External Wiki 2015-05-30T18:36:17Z

<p>Matt Lee: </p> <hr /> <div>=Precalculus=<br /> ===Good Ideas/Help===<br /> This is a section for things you want to see as part of the external wiki, Precalculus section. Provide a piped link to your page where you describe what you want to build/need help with<br /> <br /> ===In Progress===<br /> Self-explanatory <br><br /> [[Matt Lee's pages]] <br><br /> [[Power Series Diff|John's pdf Power Series with Differentiation]]<br><br /> [[Kayla's pages]]<br /> ===Completed===<br /> Once you think your page is ready to be moved to the external wiki move your link here and contact one of the Pre-calculus project members. We will let you know if it needs to be changed. Once it is approved we will let you know. Make sure you follow up to make sure you get credit for your work<br /> <br /> <br /> =Calculus=<br /> [[022 Sample Final A]]</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_5&diff=828 005 Sample Final A, Question 5 2015-05-22T04:33:17Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following inequality. Your answer should be in interval notation. <math>\frac{3x+5}{x+2}\ge 2</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |We start by subtracting 2 from each side to get <math>\frac{3x + 5}{x + 2} - \frac{2x + 4}{x + 2} = \frac{x + 1}{x + 2} \ge 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |<table border="1" cellspacing="0" cellpadding="6" align = "center"><br /> <tr><br /> <td align = "center"><math> x:</math></td><br /> <td align = "center"><math> x<-2 </math></td><br /> <td align = "center"><math> x=-2 </math></td><br /> <td align = "center"><math> -2<x<-1 </math></td><br /> <td align = "center"><math> x=-1 </math></td><br /> <td align = "center"><math>-1<x</math></td><br /> </tr><br /> <tr><br /> <td align = "center"><math> Sign: </math></td><br /> <td align = "center"><math> (+) </math></td><br /> <td align = "center"><math> VA </math></td><br /> <td align = "center"><math> (-) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (+)</math></td><br /> </tr><br /> </table><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now we just write, in interval notation, the intervals over which the denominator is nonnegative.<br /> |-<br /> | The domain of the function is: <math>(-\infty, -2) \cup [-1, \infty)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>(-\infty, -2)\cup[1, \infty)</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_2&diff=827 005 Sample Final A, Question 2 2015-05-22T04:31:10Z

<p>Matt Lee: </p> <hr /> <div>'''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><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations:<br /> |-<br /> |1) What is the domain of <math>\frac{1}{\sqrt{x}}</math>?<br /> |- <br /> |2) How can we factor <math>x^2 - x - 2</math>?<br /> |-<br /> | Answer:<br /> |- <br /> |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.<br /> |-<br /> |2) <math>x^2 - x -2 = (x - 2)(x + 1)</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |We start by factoring <math>x^2 - x - 2</math> into <math>(x - 2)(x + 1)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |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><br /> |-<br /> |<table border="1" cellspacing="0" cellpadding="6" align = "center"><br /> <tr><br /> <td align = "center"><math> x:</math></td><br /> <td align = "center"><math> x<-1 </math></td><br /> <td align = "center"><math> x=-1 </math></td><br /> <td align = "center"><math> -1<x<2 </math></td><br /> <td align = "center"><math> x=2 </math></td><br /> <td align = "center"><math>2<x</math></td><br /> </tr><br /> <tr><br /> <td align = "center"><math> Sign: </math></td><br /> <td align = "center"><math> (+) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (-) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (+)</math></td><br /> </tr><br /> </table><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now we just write, in interval notation, the intervals over which the denominator is positive.<br /> |-<br /> | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_5&diff=826 005 Sample Final A, Question 5 2015-05-22T04:28:20Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following inequality. Your answer should be in interval notation. <math>\frac{3x+5}{x+2}\ge 2</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |We start by subtracting 2 from each side to get <math>\frac{3x + 5}{x + 2} - \frac{2x + 4}{x + 2} = \frac{x + 1}{x + 2} \ge 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |<table border="1" cellspacing="0" cellpadding="6" align = "center"><br /> <tr><br /> <td align = "center"><math> x:</math></td><br /> <td align = "center"><math> x<-2 </math></td><br /> <td align = "center"><math> x=-2 </math></td><br /> <td align = "center"><math> -2<x<-1 </math></td><br /> <td align = "center"><math> x=-1 </math></td><br /> <td align = "center"><math>-1<x</math></td><br /> </tr><br /> <tr><br /> <td align = "center"><math> Sign: </math></td><br /> <td align = "center"><math> (+) </math></td><br /> <td align = "center"><math> VA </math></td><br /> <td align = "center"><math> (-) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (+)</math></td><br /> </tr><br /> </table><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>(-\infty, -2)\cup[1, \infty)</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_5&diff=825 005 Sample Final A, Question 5 2015-05-22T04:24:50Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following inequality. Your answer should be in interval notation. <math>\frac{3x+5}{x+2}\ge 2</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |We start by subtracting 2 from each side to get <math>\frac{3x + 5}{x + 2} - \frac{2x + 4}{x + 2} = \frac{x + 1}{x + 2} \ge 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |<table border="1" cellspacing="0" cellpadding="6" align = "center"><br /> <tr><br /> <td align = "center"><math> x:</math></td><br /> <td align = "center"><math> x<-1 </math></td><br /> <td align = "center"><math> x=-1 </math></td><br /> <td align = "center"><math> -1<x<2 </math></td><br /> <td align = "center"><math> x=2 </math></td><br /> <td align = "center"><math>x>2</math></td><br /> </tr><br /> <tr><br /> <td align = "center"><math> Sign: </math></td><br /> <td align = "center"><math> (+) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (-) </math></td><br /> <td align = "center"><math> 0 </math></td><br /> <td align = "center"><math> (+)</math></td><br /> </tr><br /> </table><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>(-\infty, -2)\cup[1, \infty)</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_22&diff=824 005 Sample Final A, Question 22 2015-05-22T04:23:52Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Consider the following sequence, <br><br /> <center><math> -3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots </math></center><br><br /> &nbsp;&nbsp;&nbsp;&nbsp; a. Determine a formula for <math>a_n</math>, the n-th term of the sequence. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; b. Find the sum <math> \displaystyle{\sum_{k=1}^\infty a_k}</math><br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What type of series is this?<br /> |-<br /> |2) Which formulas, about this type of series, are relevant to this question?<br /> |-<br /> |3) In the formula there are some placeholder variables. What is the value of each placeholder?<br /> |-<br /> |Answer:<br /> |-<br /> |1) This series is geometric. The giveaway is there is a number raised to the nth power.<br /> |-<br /> |2) The desired formulas are <math>a_n = a\cdot r^{n-1}</math> &nbsp; and &nbsp; <math>S_\infty = \frac{a_1}{1-r}</math><br /> |-<br /> |3) <math>a_1</math> is the first term in the series, which is <math> -3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{-1}{3}</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | The sequence is a geometric sequence. The common ratio is <math>r=\frac{-1}{3}</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | The formula for the nth term of a geometric series is <math>a_n=ar^{n-1}</math> where <math>a</math> is the first term of the sequence.<br /> |-<br /> | So, the formula for this geometric series is <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | For geometric series, <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{a}{1-r}</math> if <math>|r|<1</math>. Since <math>|r|=\frac{1}{3}</math>,<br /> |-<br /> | we have <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{-3}{1-\frac{-1}{3}}=\frac{-9}{4}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math><br /> |-<br /> |<math>\frac{-9}{4}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_22&diff=823 005 Sample Final A, Question 22 2015-05-22T04:16:36Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Consider the following sequence, <br><br /> <center><math> -3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots </math></center><br><br /> &nbsp;&nbsp;&nbsp;&nbsp; a. Determine a formula for <math>a_n</math>, the n-th term of the sequence. <br><br /> &nbsp;&nbsp;&nbsp;&nbsp; b. Find the sum <math> \displaystyle{\sum_{k=1}^\infty a_k}</math><br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What type of series is this?<br /> |-<br /> |2) Which formula, on the back page of the exam, is relevant to this question?<br /> |-<br /> |3) In the formula there are some placeholder variables. What is the value of each placeholder?<br /> |-<br /> |Answer:<br /> |-<br /> |1) This series is geometric. The giveaway is there is a number raised to the nth power.<br /> |-<br /> |2) The desired formula is <math>S_\infty = \frac{a_1}{1-r}</math><br /> |-<br /> |3) <math>a_1</math> is the first term in the series, which is <math> 5\frac{3}{5} = 3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{3}{5}</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | The sequence is a geometric sequence. The common ratio is <math>r=\frac{-1}{3}</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | The formula for the nth term of a geometric series is <math>a_n=ar^{n-1}</math> where <math>a</math> is the first term of the sequence.<br /> |-<br /> | So, the formula for this geometric series is <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | For geometric series, <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{a}{1-r}</math> if <math>|r|<1</math>. Since <math>|r|=\frac{1}{3}</math>,<br /> |-<br /> | we have <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{-3}{1-\frac{-1}{3}}=\frac{-9}{4}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math><br /> |-<br /> |<math>\frac{-9}{4}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_21&diff=822 005 Sample Final A, Question 21 2015-05-22T04:15:08Z

<p>Matt Lee: </p> <hr /> <div>''' Question''' Find the sum <br><br /> <center><math> 5 + 9 + 13 + \cdots + 49 </math></center><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) Which of the <math>S_n</math> formulas should you use?<br /> |-<br /> |2) What is the common ratio or difference?<br /> |-<br /> |3) How do you find the values you need to use the formula?<br /> |-<br /> |Answer:<br /> |-<br /> |1) The variables in the formulae give a bit of a hint. The r stands for ratio, and ratios are associated to geometric series. This sequence is arithmetic, so we want the formula that does not involve r.<br /> |-<br /> |2) Take two adjacent terms in the sequence, say <math>A_1</math> and <math>A_2</math>, and d = <math>A_2 - A_1</math><br /> |-<br /> |3) Since we have a value for d, we want to use the formula for <math>A_n</math> that involves d. <br /> |}<br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |This is the sum of an arithmetic sequence. The common difference is <math>d=4</math>. Since the formula for an arithmetic sequence is <br /> |-<br /> |<math>a_n=a_1+d(n-1)</math>, the formula for this arithmetic sequence is <math>a_n=5+4(n-1)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | We need to figure out how many terms we are adding together. To do this, we let <math>a_n=49</math> in the formula above and solve for <math>n</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | If <math>49=5+4(n-1)</math>, we have <math>44=4(n-1)</math>. Dividing by 4, we get <math>11=n-1</math>. Therefore, <math>n=12</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> |The formula for the sum of the first n terms of an arithmetic sequence is <math>S_n=\frac{1}{2}n(a_1+a_n)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> |Since we are adding 12 terms together, we want to find <math>S_{12}</math>. So, <math>S_{12}=\frac{1}{2}(12)(5+49)=324</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | 324<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_15&diff=821 005 Sample Final A, Question 15 2015-05-22T04:13:29Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Find an equivalent algebraic expression for the following, <center><math> \cos(\tan^{-1}(x))</math></center><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations<br /> |-<br /> |1) <math>\tan^{-1}(x)</math> can be thought of as <math>\tan^{-1}\left(\frac{x}{1}\right),</math> and this now refers to an angle in a triangle. What are the side lengths of this triangle?<br /> |-<br /> |Answers:<br /> |-<br /> |1) The side lengths are 1, x, and <math>\sqrt{1 + x^2}.</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |First, let <math>\theta=\tan^{-1}(x)</math>. Then, <math>\tan(\theta)=x</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Now, we draw the right triangle corresponding to <math>\theta</math>. Two of the side lengths are 1 and x and the hypotenuse has length <math>\sqrt{x^2+1}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Since <math>\cos(\theta)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}</math>, <math>\cos(\tan^{-1}(x))=\cos(\theta)=\frac{1}{\sqrt{x^2+1}}</math>.<br /> |-<br /> | <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\frac{1}{\sqrt{x^2+1}}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_14&diff=820 005 Sample Final A, Question 14 2015-05-22T04:10:06Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Prove the following identity, <br><br /> <center><math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos(\theta)}{1+\sin(\theta)}</math></center><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations: <br /> |-<br /> |1) What can you multiply <math>1- \sin(\theta)</math> by to obtain a formula that is equivalent to something involving <math>\cos</math>?<br /> |-<br /> |Answers:<br /> |-<br /> |1) You can multiply <math>1 - \sin(\theta)</math> by <math>\frac{1 + \sin(\theta)}{\1 + \sin(\theta)}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |We start with the left hand side. We have <math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{1-\sin(\theta)}{\cos(\theta)}\Bigg(\frac{1+\sin(\theta)}{1+\sin(\theta)}\Bigg)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Simplifying, we get <math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1+\sin(\theta))}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Since <math>1-\sin^2(\theta)=\cos^2(\theta)</math>, we have <br /> |-<br /> |<math>\frac{1-\sin(\theta)}{\cos(\theta)}=\frac{\cos^2(\theta)}{\cos(\theta)(1+\sin(\theta))}=\frac{\cos(\theta)}{1+\sin(\theta)}</math> <br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_13&diff=819 005 Sample Final A, Question 13 2015-05-22T03:58:30Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Give the exact value of the following if its defined, otherwise, write undefined. <br><br /> <math>(a) \sin^{-1}(2) \qquad \qquad (b) \sin\left(\frac{-32\pi}{3}\right) \qquad \qquad (c)\sec\left(\frac{-17\pi}{6}\right)</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations: <br /> |-<br /> |1) What is the domain of <math>\sin^{-1}?</math><br /> |-<br /> |2) What are the reference angles for <math>\frac{-32\pi}{3}</math> and <math>\frac{-17\pi}{6}</math>?<br /> |-<br /> |Answers:<br /> |-<br /> |1) The domain is <math>[-1, 1].</math><br /> |-<br /> |2) The reference angle for <math>\frac{-32\pi}{3}</math> is <math>\frac{4\pi}{3}</math>, and the reference angle for <math>\frac{-17\pi}{6}</math> is <math>\frac{7\pi}{6}</math><br /> |}<br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | For (a), we want an angle <math>\theta</math> such that <math>\sin(\theta)=2</math>. Since <math>-1\leq \sin (\theta)\leq 1</math>, it is impossible<br /> |-<br /> |for <math>\sin(\theta)=2</math>. So, <math>\sin^{-1}(2)</math> is undefined.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | For (b), we need to find the reference angle for <math>\frac{-32\pi}{3}</math>. If we add multiples of <math>2\pi</math> to this angle, we get the<br /> |-<br /> |reference angle <math>\frac{4\pi}{3}</math>. So, <math>\sin\left(\frac{-32\pi}{3}\right)=\sin\left(\frac{4\pi}{3}\right)=\frac{-\sqrt{3}}{2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | For (c), we need to find the reference angle for <math>\frac{-17\pi}{6}</math>. If we add multiples of <math>2\pi</math> to this angle, we get the<br /> |-<br /> |reference angle <math>\frac{7\pi}{6}</math>. Since <math>\cos\left(\frac{7\pi}{6}\right)=\frac{-\sqrt{3}}{2}</math>, we have<br /> |-<br /> |<math>\sec\left(\frac{-17\pi}{6}\right)=\sec\left(\frac{7\pi}{6}\right)=\frac{2}{-\sqrt{3}}=\frac{-2\sqrt{3}}{3}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |a) undefined<br /> |-<br /> |b) <math>\frac{-\sqrt{3}}{2}</math><br /> |-<br /> |c)<math>\frac{-2\sqrt{3}}{3}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_12&diff=818 005 Sample Final A, Question 12 2015-05-22T03:52:50Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Given that <math>\sec(\theta) = -2</math> and <math>\tan(\theta) > 0 </math>, find the exact values of the remaining trig functions.<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations<br /> |-<br /> |1) Which quadrant is <math>\theta</math> in?<br /> |-<br /> |2) Which trig functions are positive in this quadrant?<br /> |-<br /> |3) What are the side lengths of the triangle associated to <math>\theta?</math><br /> |-<br /> |Answers:<br /> |-<br /> |1) <math>\theta</math> is in the third quadrant. We know it is in the second or third quadrant since <math>\cos</math> is negative. Since \<math>\tan</math> is positive <math>\theta</math> is in the third quadrant.<br /> |-<br /> |2) <math>\tan</math> and <math>\cot</math> are both positive in this quadrant. All other trig functions are negative.<br /> |-<br /> |3) The side lengths are 2, 1, and <math>\sqrt{3}.</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | Since <math>\sec(\theta)=-2</math>, we have <math>\cos(\theta)=\frac{1}{\sec(\theta)}=\frac{-1}{2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | We look for solutions to <math>\theta</math> on the unit circle. The two angles on the unit circle with <math>\cos(\theta)=\frac{-1}{2}</math> are <math>\theta=\frac{2\pi}{3}</math> and <math>\theta=\frac{4\pi}{3}</math>.<br /> |-<br /> |But, <math>\tan\left(\frac{2\pi}{3}\right)=-\sqrt{3}</math>. Since <math>\tan(\theta)>0</math>. we must have <math>\theta=\frac{4\pi}{3}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | The remaining values of the trig functions are <br /> |-<br /> |<math>\sin(\theta)=\sin\left(\frac{4\pi}{3}\right)=\frac{-\sqrt{3}}{2}</math>,<br /> |-<br /> | <math>\tan(\theta)=\tan\left(\frac{4\pi}{3}\right)=\sqrt{3}</math><br /> |-<br /> |<math>\csc(\theta)=\csc\left(\frac{4\pi}{3}\right)=\frac{-2\sqrt{3}}{3}</math> and <br /> |-<br /> |<math>\cot(\theta)=\cot\left(\frac{4\pi}{3}\right)=\frac{\sqrt{3}}{3}</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\sin(\theta)==\frac{-\sqrt{3}}{2}</math><br /> |-<br /> |<math>\cos(\theta)=\frac{-1}{2}</math><br /> |-<br /> |<math>\tan(\theta)=\sqrt{3}</math><br /> |-<br /> |<math>\csc(\theta)=\frac{-2\sqrt{3}}{3}</math><br /> |-<br /> |<math>\cot(\theta)=\frac{\sqrt{3}}{3}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_11&diff=817 005 Sample Final A, Question 11 2015-05-22T03:43:58Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following equation in the interval <math> [0, 2\pi)</math> <br><br /> <center><math> \sin^2(\theta) - \cos^2(\theta)=1+\cos(\theta)</math></center><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: <br /> |-<br /> |1) Which trigonometric identities are useful in this problem?<br /> |-<br /> |Answer:<br /> |-<br /> |1) <math>\sin^2(\theta)=1-\cos^2(\theta)</math> and <br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | We need to get rid of the <math>\sin^2(\theta)</math> term. Since <math>\sin^2(\theta)=1-\cos^2(\theta)</math>, the equation becomes<br /> |-<br /> |<math>(1-\cos^2(\theta))-\cos^2(\theta)=1+\cos(\theta) </math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | If we simplify and move all the terms to the right hand side, we have <math>0=2\cos^2(\theta)+\cos(\theta)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now, factoring, we have <math>0=\cos(\theta)(2\cos(\theta)+1)</math>. Thus, either <math>\cos(\theta)=0</math> or <math>2\cos(\theta)+1=0</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | The solutions to <math>\cos(\theta)=0</math> in <math> [0, 2\pi)</math> are <math>\theta=\frac{\pi}{2}</math> or <math>\theta=\frac{3\pi}{2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> | The solutions to <math>2\cos(\theta)+1=0</math> are angles that satisfy <math>\cos(\theta)=\frac{-1}{2}</math>. In <math> [0, 2\pi)</math>, the <br /> |-<br /> | solutions are <math>\theta=\frac{2\pi}{3}</math> or <math>\theta=\frac{4\pi}{3}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | The solutions are <math>\frac{\pi}{2},\frac{3\pi}{2},\frac{2\pi}{3},\frac{4\pi}{3}</math>.<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_10&diff=816 005 Sample Final A, Question 10 2015-05-22T03:35:57Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) How many fractions will this decompose into? What are the denominators?<br /> |-<br /> |2) How do you solve for the numerators?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Since each of the factors are linear, and one has multipliclity 2, there will be three denominators. The linear term, x, will appear once in the denominator of the decomposition. The other two denominators will be x - 1, and <math>(x - 1)^2</math>.<br /> |-<br /> |2) After writing the equality, <math>\frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}</math>, clear the denominators, and use the cover up method to solve for A, B, and C. After you clear the denominators, the cover up method is to evaluate both sides at x = 1, 0, and any third value. Each evaluation will yield the value of one of the three unknowns.<br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Since we have a repeated factor in the denominator, we set <math>\frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get <br /> |-<br /> | <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | If we let <math>x=0</math>, we get <math>2=A</math>. If we let <math>x=1</math>, we get <math>3=C</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have<br /> |-<br /> |<math>x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x~</math>. &nbsp; So, &nbsp;<math>x+2=(2+B)x^2+(-1-B)x+2</math>. Since both sides are equal, <br /> |-<br /> |we must have <math>2+B=0</math> and <math>-1-B=1</math>. So, <math>B=2</math>. Thus, the decomposition is <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_9&diff=815 005 Sample Final A, Question 9 2015-05-22T03:29:11Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following system of equations <br><br /> ::<math> \begin{align} 2x + 3y &= & 1\\ -x + y & = & -3\end{align}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations: <br /> |-<br /> |1) What are the two methods for solving a system of equations?<br /> |-<br /> |2) How do we use the substitution method?<br /> |-<br /> |3) How do we use the elimination method?<br /> |-<br /> |Answer:<br /> |-<br /> |1) The two methods are the substitution and elimination methods.<br /> |-<br /> |2) Solve for x or y in one of the equations and substitute that value into the other equation.<br /> |-<br /> |3) Multiply one equation by some number on both sides to make one of the variables, x or y, have the same coefficient and add the equations together.<br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |Add two times the second equation to the first equation. So we are adding <math>-2x + 2y = -6</math> to the first equation.<br /> |-<br /> |This leads to:<br /> |-<br /> |<br /> ::<math>\begin{array}{rcl}<br /> 0 + 5y &=& -5\\<br /> -x + y &=& -3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |This gives us that <math>y = -1.</math><br /> |-<br /> |Now we just need to find x. So we plug in -1 for y in the second equation.<br /> |-<br /> |<br /> <math>\begin{array}{rcl}<br /> -x -1 &=& -3\\<br /> -x & =& -2\\<br /> x&=&2<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>x = 2,~ y = -1</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_8&diff=814 005 Sample Final A, Question 8 2015-05-22T03:25:16Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 3^{2x} + 3^x -2 = 0 </math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What substitution can we make to simplify the problem?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Substitute <math>y = 3^x</math> to change the original equation into <math>y^2 + y - 2 = 0</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | Start by rewriting <math>3^{2x} = \left(3^x\right)^2</math> and make the substitution <math>y = 3^x</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | After substitution we get <math>y^2 + y - 2 = (y + 2)(y - 1)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now we have to find the zeros of <math>3^x + 2 = 0</math> and <math>3^x - 1 = 0</math>. We do this by first isolating <math>3^x</math> in both equations.<br /> |-<br /> |So <math>3^x = -2</math> and <math>3^x = 1</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | We observe that <math>3^x = -2</math> has no solutions. We can solve <math>3^x = 1</math> by taking <math>log_3</math> of both sides.<br /> |-<br /> |This gives<math>\log_3\left(3^x\right) = x = \log_3(1) = 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | x = 0<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_8&diff=813 005 Sample Final A, Question 8 2015-05-22T03:24:51Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 3^{2x} + 3^x -2 = 0 </math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What substitution can we make to simplify the problem?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Substitute y = <math>3^x</math> to change the original equation into <math>y^2 + y - 2 = 0</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | Start by rewriting <math>3^{2x} = \left(3^x\right)^2</math> and make the substitution <math>y = 3^x</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | After substitution we get <math>y^2 + y - 2 = (y + 2)(y - 1)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now we have to find the zeros of <math>3^x + 2 = 0</math> and <math>3^x - 1 = 0</math>. We do this by first isolating <math>3^x</math> in both equations.<br /> |-<br /> |So <math>3^x = -2</math> and <math>3^x = 1</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | We observe that <math>3^x = -2</math> has no solutions. We can solve <math>3^x = 1</math> by taking <math>log_3</math> of both sides.<br /> |-<br /> |This gives<math>\log_3\left(3^x\right) = x = \log_3(1) = 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | x = 0<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_8&diff=812 005 Sample Final A, Question 8 2015-05-22T03:24:37Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 3^{2x} + 3^x -2 = 0 </math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What substitution can we make to simplify the problem?<br /> |-<br /> |Answer:<br /> |-<br /> |1) Substitute y = <math>3^x</math> to change the original equation into <math>y^2 + y - 2 = 0</math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | Start by rewriting <math>3^{2x} = \left(3^x\right)^2</math> and make the substitution <math>y = 3^x</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | After substitution we get <math>y^2 + y - 2 = (y + 2)(y - 1)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Now we have to find the zeros of <math>3^x + 2 = 0</math> and <math>3^x - 1 = 0</math>. We do this by first isolating <math>3^x</math> in both equations.<br /> |-<br /> |So <math>3^x = -2</math> and <math>3^x = 1</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | We observe that <math>3^x = -2</math> has no solutions. We can solve <math>3^x = 1</math> by taking <math>log_3</math> of both sides.<br /> |-<br /> |This gives<math>\log_3\left(3^x\right) = x = \log_3(1) = 0</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | x = 0<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_7&diff=811 005 Sample Final A, Question 7 2015-05-22T03:22:40Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following equation, &nbsp;&nbsp;&nbsp;&nbsp; <math> 2\log_5(x) = 3\log_5(4)</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What logarithm rule is relevant for dealing with the coefficients of the logarithms?<br /> |-<br /> |2) How do we remove the logs?<br /> |-<br /> |Answer:<br /> |-<br /> |1) One of the rules of logarithms says that <math> r\log(x) = \log(x^r)</math><br /> |-<br /> |2) The definition of logarithm tells us that if <math> \log_5(x) = y </math>, then <math> 5^y = x </math><br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1<br /> |-<br /> | Use the rules of logarithms to move the 2 and the 3 to exponents. So <math>\log_5(x^2) = \log_5(4^3)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2<br /> |-<br /> | By the definition of logarithm, we find that <math>x^2 = 4^3</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3<br /> |-<br /> | Taking the square root of both sides we get <math>x = 8</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer<br /> |-<br /> | <math> x = 8</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_6&diff=810 005 Sample Final A, Question 6 2015-05-22T03:19:33Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Factor the following polynomial completely, &nbsp;&nbsp;&nbsp;&nbsp;<math>p(x) = x^4 + x^3 + 2x-4 </math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> !Foundations<br /> |-<br /> |1) What does the Rational Zeros Theorem say about the possible zeros?<br /> |-<br /> |2) How do you check if a possible zero is actually a zero?<br /> |-<br /> |3) How do you find the rest of the zeros?<br /> |-<br /> |Answer:<br /> |-<br /> |1) The possible divisors can be found by finding the factors of -10, in a list, and the factors of 1, in a second list. Then write down all the fractions with numerators from the first list and denominators from the second list.<br /> |-<br /> |2) Use synthetic division, or plug a possible zero into the function. If you get 0, you have found a zero. <br /> |-<br /> |3) After your reduce the polynomial with synthetic division, try and find another zero from the list you made in part a). Once you reach a degree 2 polynomial you can finish the problem with the quadratic formula.<br /> |}<br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | First, we use the Rational Zeros Theorem to note that the possible zeros are: <math>\{\pm 1, \pm 2, \pm 4 \}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Now we start checking which of the possible roots are actually roots. We find that 1 is a zero, and apply either synthetic division or long division to get <math>x^4 + x^3 +2x - 4 = (x - 1)(x^3 +2x^2 + 2x +4)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | We continue checking zeros and find that -2 is a zero. Applying synthetic division or long division we can simplify the polynomial down to:<br /> |-<br /> |<math>x^4 +x^3+ 2x -4 = (x - 1)(x + 2)(x^2 + 2)</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | Now we can finish the problem by applying the quadratic formula or just finding the roots of <math>x^2 + 2</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>x^4 + x^3 +2x - 4 = (x - 1)(x + 2)(x - \sqrt{2}i)(x + \sqrt{2}i)</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_4&diff=809 005 Sample Final A, Question 4 2015-05-22T03:16:44Z

<p>Matt Lee: </p> <hr /> <div>'''Question''' Find the inverse of the following function <math> f(x) = \frac{3x}{2x-1}</math><br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations: <br /> |-<br /> |1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>?<br /> |-<br /> |Answer:<br /> |-<br /> |1) you would replace f(x) by y, switch x and y, and finally solve for y.<br /> |}<br /> <br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | Switch f(x) for y, to get <math>y = \frac{3x}{2x-1}</math>, then switch y and x to get <math>x = \frac{3y}{2y-1}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Now we have to solve for y:<br /> ::<math> \begin{array}{rcl}<br /> x & = & \frac{3y}{2y-1}\\<br /> x(2y - 1) & = & 3y\\<br /> 2xy - x & = & 3y\\<br /> 2xy - 3y & = & x\\<br /> y(2x - 3) & = & x\\<br /> y & = & \frac{x}{2x - 3}<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math> y = \frac{x}{2x-3}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_3&diff=808 005 Sample Final A, Question 3 2015-05-22T03:12:46Z

<p>Matt Lee: </p> <hr /> <div>'''Question ''' Find f <math>\circ</math> g and its domain if <math>f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations: <br /> |-<br /> |1) How do you compose two functions, such as given <math>f = x^2</math>&nbsp; and &nbsp; <math>g = x + 1</math>, what is f<math>\circ</math>g?<br /> |-<br /> |2) When should a point x be in the domain of f<math>\circ</math>g?<br /> |-<br /> |Answers:<br /> |-<br /> |1) We replace all occurrences of x in f with g, so <math>f\circ g = (x + 1)^2</math>.<br /> |-<br /> |2) A point should be in the domain of f<math>\circ</math>g when it is in the domain of g, and g(x) is in the domain of f.<br /> |}<br /> <br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1: <br /> |-<br /> |First we find the domain of g. Since f <math>\circ</math> g = f(g(x)). So if x is not in the domain of g, it is not in the domain of f <math>\circ</math> g. The domain of g is <math>[1, \infty)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2: <br /> |-<br /> |To find f <math>\circ</math> g we replace any occurrence of x in f with g, to yield <math>(\sqrt{x - 1})^2 + 1 = x - 1 + 1 = x </math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answers: <br /> |-<br /> |f <math>\circ</math> g = <math> x </math>, and the domain is <math>[1, \infty)</math>.<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_3&diff=807 005 Sample Final A, Question 3 2015-05-22T03:12:34Z

<p>Matt Lee: </p> <hr /> <div>'''Question ''' Find f <math>\circ</math> g and its domain if <math>f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Foundations: <br /> |-<br /> |1) How do you compose two functions, such as given <math>f = x^2</math>&nbsp; and &nbsp; <math>g = x + 1</math>, what is f<math>\circ</math>g?<br /> |-<br /> |2) When should a point x be in the domain of f<math>\circ</math>g?<br /> |-<br /> |Answers:<br /> |-<br /> |1) We replace all occurrences of x in f with g, so <math>f\circ g = (x + 1)^2</math>.<br /> |-<br /> |2) A point should be in the domain of f<math>\circ</math>g when it is in the domain of g, and g(x) is in the domain of f.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1: <br /> |-<br /> |First we find the domain of g. Since f <math>\circ</math> g = f(g(x)). So if x is not in the domain of g, it is not in the domain of f <math>\circ</math> g. The domain of g is <math>[1, \infty)</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2: <br /> |-<br /> |To find f <math>\circ</math> g we replace any occurrence of x in f with g, to yield <math>(\sqrt{x - 1})^2 + 1 = x - 1 + 1 = x </math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answers: <br /> |-<br /> |f <math>\circ</math> g = <math> x </math>, and the domain is <math>[1, \infty)</math>.<br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_10&diff=793 005 Sample Final A, Question 10 2015-05-20T05:47:54Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Since we have a repeated factor in the denominator, we set <math>\frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get <br /> |-<br /> | <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | If we let <math>x=0</math>, we get <math>2=A</math>. If we let <math>x=1</math>, we get <math>3=C</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have<br /> |-<br /> |<math>x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x~</math>. &nbsp; So, &nbsp;<math>x+2=(2+B)x^2+(-1-B)x+2</math>. Since both sides are equal, <br /> |-<br /> |we must have <math>2+B=0</math> and <math>-1-B=1</math>. So, <math>B=2</math>. Thus, the decomposition is <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_10&diff=792 005 Sample Final A, Question 10 2015-05-20T05:47:25Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Since we have a repeated factor in the denominator, we set <math>\frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get <br /> |-<br /> | <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | If we let <math>x=0</math>, we get <math>2=A</math>. If we let <math>x=1</math>, we get <math>3=C</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have<br /> |-<br /> |<math>x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x~</math>. So, <math>~x+2=(2+B)x^2+(-1-B)x+2</math>. Since both sides are equal, <br /> |-<br /> |we must have <math>2+B=0</math> and <math>-1-B=1</math>. So, <math>B=2</math>. Thus, the decomposition is <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_10&diff=791 005 Sample Final A, Question 10 2015-05-20T05:46:56Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> | First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> | Since we have a repeated factor in the denominator, we set <math>\frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 3:<br /> |-<br /> | Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get <br /> |-<br /> | <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>. <br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 4:<br /> |-<br /> | If we let <math>x=0</math>, we get <math>2=A</math>. If we let <math>x=1</math>, we get <math>3=C</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 5:<br /> |-<br /> | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have<br /> |-<br /> |<math>x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x</math>. So, <math>x+2=(2+B)x^2+(-1-B)x+2</math>. Since both sides are equal, <br /> |-<br /> |we must have <math>2+B=0</math> and <math>-1-B=1</math>. So, <math>B=2</math>. Thus, the decomposition is <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math>.<br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> | <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_9&diff=788 005 Sample Final A, Question 9 2015-05-19T06:48:53Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following system of equations <br><br /> ::<math> \begin{align} 2x + 3y &= & 1\\ -x + y & = & -3\end{align}</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |Add two times the second equation to the first equation. So we are adding <math>-2x + 2y = -6</math> to the first equation.<br /> |-<br /> |This leads to:<br /> |-<br /> |<br /> ::<math>\begin{array}{rcl}<br /> 0 + 5y &=& -5\\<br /> -x + y &=& -3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |This gives us that <math>y = -1.</math><br /> |-<br /> |Now we just need to find x. So we plug in -1 for y in the second equation.<br /> |-<br /> |<br /> <math>\begin{array}{rcl}<br /> -x -1 &=& -3\\<br /> -x & =& -2\\<br /> x&=&2<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>x = 2,~ y = -1</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_9&diff=787 005 Sample Final A, Question 9 2015-05-19T06:47:38Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following system of equations <br><br /> ::<math> \begin{align} 2x + 3y &= & 1\\ -x + y & = & -3\end{align}</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |Add two times the second equation to the first equation. So we are adding <math>-2x + 2y = -6</math> to the first equation.<br /> |-<br /> |This leads to:<br /> |-<br /> |<br /> ::<math>\begin{array}{rcl}<br /> 0 + 5y &=& -5\\<br /> -x + y &=& -3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |This gives us that <math>y = -1.</math><br /> |-<br /> |Now we just need to find x. So we plug in -1 for y in the second equation.<br /> |-<br /> |<br /> <math>\begin{array}{rcl}<br /> -x -1 &=& -4\\<br /> -x & =& -3\\<br /> x&=&3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>x = 3,~ y = -1</math><br /> |}</div>

Matt Lee https://gradwiki.math.ucr.edu/index.php?title=005_Sample_Final_A,_Question_9&diff=786 005 Sample Final A, Question 9 2015-05-19T06:47:25Z

<p>Matt Lee: </p> <hr /> <div>''' Question ''' Solve the following system of equations <br><br /> ::<math> \begin{align} 2x + 3y &= & 1\\ -x + y & = & -3\end{align}</math><br /> <br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 1:<br /> |-<br /> |Add two times the second equation to the first equation. So we are adding <math>-2x + 2y = -6</math> to the first equation.<br /> |-<br /> |This leads to:<br /> |-<br /> |<br /> ::<math>\begin{array}{rcl}<br /> 0 + 5y &=& -5\\<br /> -x + y &=& -3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Step 2:<br /> |-<br /> |This gives us that <math>y = -1.</math><br /> |-<br /> |Now we just need to find x. So we plug in -1 for y in the second equation.<br /> |-<br /> |<br /> <math>\begin{array}{rcl}<br /> -x -1 &=& -4\\<br /> -x & =& -3\\<br /> x&=&3<br /> \end{array}</math><br /> |}<br /> <br /> {| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br /> ! Final Answer:<br /> |-<br /> |<math>x = 3, y = -1</math><br /> |}</div>

Matt Lee