005 Sample Final A, Question 10 - Revision history

Matt Lee at 03:35, 22 May 2015

2015-05-22T03:35:57Z

Matt Lee at 05:47, 20 May 2015

2015-05-20T05:47:54Z

Matt Lee at 05:47, 20 May 2015

2015-05-20T05:47:25Z

Matt Lee at 05:46, 20 May 2015

2015-05-20T05:46:56Z

Kayla Murray at 22:35, 19 May 2015

2015-05-19T22:35:57Z

Matt Lee: Created page with "''' Question ''' Write the partial fraction decomposition of the following, $\frac{x+2}{x^3-2x^2+x}$ {| class="mw-collapsible mw-collapsed" sty..."

2015-05-01T04:22:38Z

Created page with "''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center> {| class="mw-collapsible mw-collapsed" sty..."

New page

''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
|-
|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

← Older revision		Revision as of 03:35, 22 May 2015
Line 1:		Line 1:
	''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center>		''' Question ''' Write the partial fraction decomposition of the following, <center> <math> \frac{x+2}{x^3-2x^2+x}</math></center>
		+
		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Foundations
		+	\|-
		+	\|1) How many fractions will this decompose into? What are the denominators?
		+	\|-
		+	\|2) How do you solve for the numerators?
		+	\|-
		+	\|Answer:
		+	\|-
		+	\|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>.
		+	\|-
		+	\|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.
		+	\|}

@@ Line 33: / Line 33: @@
 | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have
 |-
-|<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,
+|<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,
 |-
 |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>.

@@ Line 33: / Line 33: @@
 | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have
 |-
-|<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,
+|<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,
 |-
 |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>.

@@ Line 33: / Line 33: @@
 | To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have
 |-
-|<math>x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x. So, <math>x+2=(2+B)x^2+(-1-B)x+2</math>. Since both sides are equal,
+|<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,
 |-
 |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>.

@@ Line 3: / Line 3: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Step 1:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+| First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math>
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+| 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>.
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+| Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+| <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>.
 |-
-|e) True.
+| 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>.
 |-
-|f) False.
+| To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have
 |}

005 Sample Final A, Question 10 - Revision history

Matt Lee at 03:35, 22 May 2015

Matt Lee at 05:47, 20 May 2015

Matt Lee at 05:47, 20 May 2015

Matt Lee at 05:46, 20 May 2015

Kayla Murray at 22:35, 19 May 2015

Matt Lee: Created page with "''' Question ''' Write the partial fraction decomposition of the following, \frac{x+2}{x^3-2x^2+x} {| class="mw-collapsible mw-collapsed" sty..."

Matt Lee: Created page with "''' Question ''' Write the partial fraction decomposition of the following, $\frac{x+2}{x^3-2x^2+x}$ {| class="mw-collapsible mw-collapsed" sty..."