Difference between revisions of "005 Sample Final A, Question 10"

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(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...")
 
 
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''' 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>
  
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!Foundations
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|1) How many fractions will this decompose into? What are the denominators?
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|-
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|2) How do you solve for the numerators?
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|Answer:
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|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>.
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|-
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|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.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 1:
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|-
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| First, we factor the denominator. We have <math>x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2</math>
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|}
  
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Final Answers
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! Step 2:
 
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|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>
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| 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>.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 3:
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|-
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| Multiplying both sides of the equation by the denominator <math>x(x-1)^2</math>, we get
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|-
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| <math>x+2=A(x-1)^2+B(x)(x-1)+Cx</math>.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 4:
 
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|-
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|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.
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| 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>.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Step 5:
 
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|-
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|c) False. <math>y = x^2</math> does not have an inverse.
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| To solve for <math>B</math>, we plug in <math>A=2</math> and <math>C=3</math> and simplify. We have
 
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|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
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|<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,
 
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|e) True.
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|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>.
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Final Answer:
 
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|f) False.
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| <math>\frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}</math>
 
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Latest revision as of 19:35, 21 May 2015

Question Write the partial fraction decomposition of the following,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x+2}{x^3-2x^2+x}}
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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - 1)^2} .
2) After writing the equality, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}} , 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.


Step 1:
First, we factor the denominator. We have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2}
Step 2:
Since we have a repeated factor in the denominator, we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}} .
Step 3:
Multiplying both sides of the equation by the denominator Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(x-1)^2} , we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+2=A(x-1)^2+B(x)(x-1)+Cx} .
Step 4:
If we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2=A} . If we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3=C} .
Step 5:
To solve for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} , we plug in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C=3} and simplify. We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+2=2(x-1)^2+B(x)(x-1)+3x=2x^2-4x+2+Bx^2-Bx+3x~} .   So,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+2=(2+B)x^2+(-1-B)x+2} . Since both sides are equal,
we must have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2+B=0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1-B=1} . So, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=2} . Thus, the decomposition is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}} .
Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x+2}{x(x-1)^2}=\frac{2}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}}
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