Chapter 3 – Polynomial and Rational Functions 3.7 - Rational Functions.
Chapter 3 Section 3.5 Rational Functions and Models.
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Chapter 3 Section 3.5 Rational Functions and Models
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Transcript of Chapter 3 Section 3.5 Rational Functions and Models.
Chapter 3
Section 3.5
Rational Functions and Models
11/11/2012 Section 3.4 22
Definition
f(x) is a rational function if and only if f(x) =
Examples 1. f(x) =
2. f(x) =
3. f(x) =
Rational Functions
p(x)q(x)
where p(x) and q(x) are polynomial functions with q(x) 0
3x2 + 4x + 1
x3 – 1
3x2 – 27
x – 3
x3/2 – 8
x2 + 1
=3(x – 3)(x + 3)
x – 3 = 3x + 9 , for x ≠ 3
Question: Is this a rational function ?
11/11/2012 Section 3.4 33
Domains
Where are rational functions defined?
Examples 1. f(x) =
2. f(x) =
3. f(x) =
Rational Functions
3x2 + 4x + 1
x3 – 1
3x2 – 27
x – 3
Dom f(x) = { x | x ≠ 1 }
Dom f(x) = { x | x ≠ 3 }
6x2 – x – 2 x2 + x – 6
Dom f(x) = { x | x ≠ – 3, x ≠ 2 }
3x2 + 4x + 1
(x – 1)(x2 + x + 1)=
6x2 – x – 2 (x + 3)(x – 2)
=
11/11/2012 Section 3.4 4
Asymptotes Asymptotes are lines that the graph of a function f(x)
approaches closely as x approaches some k or ±
Vertical Asymptote: line x = k such that either f(x)
Horizontal Asymptote: line y = k such that f(x)
as x
4
Rational Functions
∞or f(x) as x approaches k∞–
k
∞ or x ∞–
∞
Question: Can the graph of f(x) cross its asymptotes ?
11/11/2012 Section 3.4 55
Asymptote Examples
1. f(x) =
2. f(x) =
Rational Functions
1x x
y
x
y
1 x + 1
x = –1
x = 0
y = 0
11/11/2012 Section 3.4 66
Asymptote Examples
3. f(x) =
4. f(x) =
Rational Functions
x
y
x
y
y = 1
x = 5
x2 – 4 x – 2
2●
x + 1x – 5
11/11/2012 Section 3.4 77
Asymptote Examples
5. f(x) =
Sketch asymptotes, intercepts and the graph
Rational Functions
x
y
5/2●
4x – 102x – 5
4x – 102x – 5
f(x) =2(2x – 5)
2x – 5=
= 2… provided x ≠ 5/2
●(0, 2)
No asymptotes !
One intercept: (0, 2)
Function is linear, but undefined at x = 5/2
Domain ?
Range ?
Domain = { x │ x ≠ 5/2 }= ( – , 5/2 ) ( 5/2 , )∞ ∞
Range = { 2 }
11/11/2012 Section 3.4 88
Asymptote Examples
6. f(x) =
f(1) does not exist since
x3 – 1 = 0 when x = 1
Rational Functions
x
y
Line y = 0
Line x = 1
11x – 2x3 – 1
Horizontal asymptoteas x ∞±Vertical asymptote
0f(x)
Question: Does the graph cross an asymptote ? YES !
f(x) = 0 atx = 2 11
so the graph cuts the asymptote y = 0 at , 02 11
)(
11/11/2012 Section 3.4 9
Asymptote Review Vertical Asymptote: line x = k such that either f(x)
Horizontal Asymptote: line y = k such that f(x)
as x
9
Rational Functions
∞or f(x) as x approaches k∞–
k
∞ or x ∞–
Question: Can the graph of f(x) cross its asymptote ?
Consider f(x) =5x2 + 8x – 3
3x2 + 2 and g(x) =
11x + 2
2x3 – 1 ... and their horizontal asymptotesWhat about vertical asymptotes ?
Question: Are asymptotes always vertical or horizontal ?
11/11/2012 Section 3.4 1010
Asymptote Examples
7. f(x) =
Since 3x2 + 2 is never zero
for any real x there is no
vertical asymptote
Rewriting
Thus f(x) has a horizontal asymptote of y =
Rational Functions
x
y
5x2 + 8x – 33x2 + 2
Horizontal asymptote
as x ∞±
5 x8 –
x2
3+
3x2
2+
f(x) = 35
53
y = 35
Question: Where does the graph cross its asymptote ?
y = f(x)
●
3519
24 ( ),
11/11/2012 Section 3.4 1111
Asymptote Examples
8. f(x) =
Rational Functions
x
y
y = x + 3
x = 1
x2 + 2x + 1x – 1
Oblique (slant) Asymptote
Oblique Asymptote: line y = ax + b such that f(x) y
as x ∞±
Occurs when deg (numerator) = deg (denominator) + 1
By synthetic division ...
1 1 2 1
113
34
f(x) = x + 3 +4
x – 1
Thus
f has a vertical asymptote at x = 1and
f(x) y = x + 3 as x ±
VerticalAsymptote
11/11/2012 Section 3.4 1212
Finding Vertical and Horizontal Asymptotes
Vertical Asymptotes for f(x) = Find values of x , say x = k, where q(x) = 0
f(x) fails to exist at x = k
… BUT might not have an asymptote there
if (x – k) is also a factor of p(x) Ensure f(x) is reduced to lowest terms Check x = 0 for vertical intercept
Horizontal and Oblique Asymptotes for f(x) =
Determine what f(x) approaches as x approaches
Check f(x) = 0 for horizontal intercept
Rational Functions
p(x)q(x)
p(x)
q(x)
∞±
11/11/2012 Section 3.4 1313
Find Horizontal / Oblique Asymptotes
9. f(x) =
10. f(x) =
11. f(x) =
Rational Functions
4x3 – 2 x + 2
6x2 – x – 2 2x2 – 3x + 4
x2 + 2x + 1 x – 1
11/11/2012 Section 3.4 1414
Rational Functions for “Large” x
For the function
Rational Functions
R(x) =3x + 1
x – 2 fill in the table
3 10 100 1,000 10,000 100,000 1,000,000 x
3x + 1
x – 2
R(x)
10 31 301 3,001 30,001 300,001 3,000,001
1 8 98 998 9,998 99,998 999,998
10 3.9 3.07 3.007 3.0007 3.00007 3.000007
Thus as x
1,000,000 we see that3x + 1 3,000,001 ≈ 3,000,000 = 3x
x – 2 999,998 ≈ 1,000,000 = x
R(x) 3.000007 ≈ 3 = (3x)/x
11/11/2012 Section 3.4 1515
How do we solve equations of form:
Method 1: Clear Fractions
Examples:
1. Solve:
Solving Rational Equations
h(x) = g(x)
f(x)
= 1523
x + 2
= 1523
x + 2· (x + 2)· (x + 2)
23 = 15x + 30
–7 = 15x
=–7
15x
Solution Set: { }7
15–
11/11/2012 Section 3.4 1616
Method 1: Clear Fractions
2. Solve:
Solving Rational Equations
=x + 1x + 2
x + 5x + 7
· (x + 2)(x + 7)
=x + 1x + 2
x + 5x + 7· (x + 2)(x +
7)(x + 5)(x + 2)=(x + 1)(x + 7)
x2 + 7x + 10=x2 + 8x + 7
7x + 10=8x + 7
3=x
Solution Set: { 3 }
11/11/2012 Section 3.4 1717
Method 2: Cross Multiplication
Basic Principle:
Examples:
1. Solve:
Solving Rational Equations
=x + 1x + 2
x + 5x + 7
(x + 2)(x + 5)=(x + 1)(x + 7)
x2 + 7x + 10=x2 + 8x + 7
7x + 10=8x + 7
3=x
=ab
cd
if and only if ad = bc
Solution Set: { 3 }
11/11/2012 Section 3.4 1818
Method 2: Cross Multiplication
2. Solve:
Cross multiplying
Solving Rational Equations
=x – 3
71
x + 3
7=(x – 3)(x + 3)
7=x2 – 9
0=x2 – 16
0=(x + 4)(x – 4)
0=x + 4 0=x – 4 OR
– 4=x 4=x
=x2à 16ñ
Solution Set: { – 4, 4 }
=x ± 4
16=x2
Factoring and Zero Product Property Square Root Property
11/11/2012 Section 3.4 1919
Method 2: Cross Multiplication
3. Solve:
Cross multiplying
Solving Rational Equations
=x + 15x – 3
23
2(5x – 3)=3(x + 1)
10x – 6=3x + 3
7x=9
=97
x
Solution Set: { }9
7
11/11/2012 Section 3.4 2020
Method 3: Graphical Approach
1. Solve:
Let y1 =
Solving Rational Equations
=x + 1x – 5
2
x + 1x – 5
and y2 = 2
x
y
3
3 6 9 11–2
–3
y1
Vertical Asymptotex = 5
y2 Horizontal Asymptote
y = 1
Intersection at (11, 2)
(11, 2)
Hence: x = 11
So y1 = y2
For what x is this true ?
Intercepts for y1 :Horizontal : ( –1, 0 ) Vertical : ( 0, –1/5 )
11/11/2012 Section 3.4 21
Direct Variation Output varies directly with input
Example: y = kx
Inverse Variation Output varies inversely with input
Example: y = kx–1
Inverse Variation Functions Output varies inversely with xn
Example: y = kx–n
21
Direct and Inverse Variation
k is the constant of variation
yx k=OR
yx = kOR
yxn = kOR
11/11/2012 Section 3.4 2222
Direct Variation The resultant force acting on an object of mass m is
directly proportional to the acceleration of the object
F = ma
F varies directly with a -- Newton’s Second Law Constant of variation is m
Variation Examples
OR
= m F a
11/11/2012 Section 3.4 2323
Direct Variation Example Weight and Mass
Excluding other external forces, the only force acting on an object of mass m is the force of gravity mg, where g is the acceleration due to gravity
To lift the object, the force of gravity must be overcome This force is called weight and given by F = W = mg Clearly weight varies directly with mass
Variation Examples
; that is
F = mg
11/11/2012 Section 3.4 2424
Inverse Variation At constant temperature the volume of n moles of gas
is inversely proportional to the pressure of the gas
PV = nRT OR P = (nRT)V–1 P varies inversely with V -- Ideal Gas Law Constant of variation is nRT
Variation Examples
11/11/2012 Section 3.4 2525
Function of Variation The earth’s gravitational force acting on an object of
mass m is inversely proportional to the square of the distance between the mass and the center of the earth
F
F varies inversely with r2 -- Law of Gravity
Constant of variation is GMm, where G is the earth’s gravitational constant, M is the mass of the earth, and m is the mass of the object
This is just one of many inverse square laws
Variation Examples
=GMm
r2 OR Fr2 = GMm
11/11/2012 Section 3.4 2626
Think about it !
