Notes 7.4 – Partial Fractions · Notes 7.4 – Partial Fractions. I. Composing Fractions A.) Ex....
Transcript of Notes 7.4 – Partial Fractions · Notes 7.4 – Partial Fractions. I. Composing Fractions A.) Ex....
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Notes 7.4 – Partial Fractions
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I. Composing FractionsA.) Ex. 1– Rewrite the following expression as a
single rational expression.
3 24 3x x
2
3( 3) 2( 4) 5 1( 4)( 3) 12x x xx x x x
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II. Decomposing FractionsA.) Steps:
1.) If the degree of f ≥ the degree of d, then
2.) Factor d(x) into a product of factors of the form (mx + n)u or (ax2 + bx + c)v when (ax2 + bx + c) is irreducible.
( ) ( )( )( ) ( )
f x r xq xd x d x
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3a.) If d(x) is of the form (mx + n)u
1 2
2 ... uu
AA Amx n mx n mx n
1 1 2 2
22 2 2... v v
v
B x CB x C B x Cax bx c ax bx c ax bx c
3b.) If d(x) is of the form (ax2 + bx + c)v
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22 2
2 5
2 3 1
x
x x x x
1 2 1 1 1 2 2
22 2 22 3 1 3 1
A A B C x D C x Dx x x x x x x
B.) For example,
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1 22 5
3 2 3 2A Ax
x x x x
2 53 2x
x x
C.) Ex. 2- Find the partial fraction decomposition of
Multiply both sides by the LCD
1 22 53 2
3 2 3 2A Axx x
x x x x
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1 22 5 2 3x A x A x
1 2 1 22 5 2 3x A x A x A A 1 1 2 22 5 2 3x A x A A x A
1 2 1 22 5 2 3x x A A A A
1 2
1 2
25 2 3
A AA A
Therefore
1
2
1 1 22 3 5
AA
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1 92 5 5 53 2 3 2x
x x x x
2 5 1 93 2 5 3 5 2x
x x x x
1
2
1595
AA
Solving the matrix equation gives us
SUPPORT GRAPHICALLY!!!
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1 22 5 2 3x A x A x
1 22 2 5 2 2 2 3A A
What if we were to substitute x = 2 into the equation below?
2
2
9 0 595
A
A
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1 22 3 5 3 2 3 3A A
1 21 5 0A A
Now, substitute x = -3 into the equation.
115
A
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2
3 2
2 46 9
x xx x x
31 2
23 3AA A
x x x
D.) Ex.3 - Find the partial fraction decomposition of
2
22 4
3x xx x
221 2 32 4 3 3x x A x A x x A x
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2 2 21 1 1 2 2 32 4 6 9 3x x A x A x A A x A x A x
21 2 1 2 3 16 3 9A A x A A A x A
2 21 2 1 2 3 16 3 9A x A x A x A x A x A
1
2
3
1 1 0 16 3 1 29 0 0 4
AAA
1
2
3
495973
AAA
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221 2 32 4 3 3x x A x A x x A x
21 2 34 3 0 3 0A A A
Try this, substitute x = 0 into the equation below.
1
1
4 949
A
A
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221 2 32 4 3 3x x A x A x x A x
21 2 37 0 3 0 3A A A
Now, substitute x = -3 into the equation below.
3
3
7 373
A
A
We can now choose any value for x and substitute it into one of the equations along with A1 and A3and solve for A2.
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III. Denominators of Irreducible Quadratic Factors
1 1 2 2
22 2 2... v v
v
B x CB x C B x Cax bx c ax bx c ax bx c
A.) If d(x) is of the form (ax2 + bx + c)v
Since subscripts can get confusing, we can simply use different letters.
22 2 2... v
Ax B Cx D Yx Zax bx c ax bx c ax bx c
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B.) Ex. 4 – Write the partial fraction decomposition of
3 2
22 4
x x
x
3 2
2 222 244 4
x x Ax B Cx Dxx x
3 2 2 4x x x Ax B Cx D
3 2222 222 2
444 4
x x Ax B Cx Dxxx x
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3 2 3 2 4 4x x Ax Bx A C x B D
3 2 3 2 4 4x x Ax Bx Ax B Cx D
4 1 04C
C
11
4 04 0
AB
A CB D
4(1) 04D
D
3 2
2 222 2
1 4( 1)44 4
x x x xxx x
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C.) Ex. 5 – Find the partial fraction decomposition of
Unfortunately, there is no shortcut method for denominators with irreducible quadratics.
2
3 2
4 11
x xx x x
2 2 2
3 2 2 2
4 1 4 1 4 11 1 1 1 1
x x x x x xx x x x x x x x
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2
22
4 11 11 1
x x A Bx Cx xx x
2 24 1 1 1x x A x Bx C x
2
222
4 11 11 11 1
x x A Bx Cx xx xx x
2 24 1x x x A B x C B A C
2 2 24 1x x Ax A Bx Cx Bx C
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1 1 0 10 1 1 41 0 1 1
141
A BC BA C
32
2
ABC
1 0 0 30 1 0 20 0 1 2
A
rref(A)
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2
3 2 2
4 1 3 2 21 1 1
x x xx x x x x
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Before we graph it, what can you tell me about what the graph will look like?
IV. Applications of Partial Fractions
2
2
2 154
x xx
A.) Ex. 6 - Graph the following function on your calculator.
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2
2
2 15 5 924 4 2 4 2
x xx x x
Now, lets look at the partial fraction decomposition of the function and the graphs of the individual pieces.
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2
2
2 15 5 924 4 2 4 2
x xx x x
2y First Piece - Models the end
behavior of the graph
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5
4 2y
x
2
2
2 15 5 924 4 2 4 2
x xx x x
2nd Piece - Models the behavior of the graph near the vertical asymptote x =-2
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9
4 2y
x
2
2
2 15 5 924 4 2 4 2
x xx x x
3rd Piece - Models the behavior of the graph near the vertical asymptote x =2
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2
2
2 15 5 924 4 2 4 2
x xx x x
Combining the 3 pieces gives us this…
Or, more precisely, something like this…
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B.) Ex. 7 – Explain in sentences the three different parts to the graph of the following function. Then, graph the function .
1 5( ) 13 2
y f x xx x
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Second, the graph of the function at x = -3 will closely resemble the behavior of the graph of the hyperbola with its branches in the upper right and lower left quadrants.
13
yx
First, the graph’s end-behavior asymptote will be modeled by the graph y = -x +1.
Third, the graph of the function at x = 2 will closely resemble the behavior of the graph of the hyperbola with its branches in the upper left and lower right quadrants.
52
yx
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Combining the three separate parts gives us a graph that looks like