Transcript No Slide Title

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Nonlinear Functions
and Equations
Nonlinear Functions and Their graphs
Polynomial Functions and Models
Real Zeros of Polynomial Functions
The Fundamental Theorem of Algebra
Rational Functions and Models
Polynomial and Rational Inequalities
Power Functions and Radical Equations
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4.1
Nonlinear Functions
and Their Graphs
♦ Learn terminology about polynomial functions
♦ Identify intervals where a function is increasing or
decreasing
♦ Find extrema of a function
♦ Identify symmetry in a graph of a function
♦ Determine if a function is odd, even, or neither
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Polynomial Functions
Polynomial functions are frequently used to
approximate data.
A polynomial function of degree 2 or higher is a
nonlinear function.
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Slide 4- 4
Increasing or Decreasing Functions
The concept of increasing and decreasing relate to
whether the graph of a function rises or falls.
• Moving from left to right along a graph of an
increasing function would be uphill.
• Moving from left to right along a graph of a
decreasing function would be downhill.
We speak of a function f
increasing or decreasing
over an interval of its domain.
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Slide 4- 5
Increasing or Decreasing Functions continued
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Slide 4- 6
Example
Use the graph of f ( x )  3x 3  9x 2  4 shown below and
interval notation to identify where f is increasing or
decreasing.
Solution
Moving from left to right on the
graph of f, the y-values
decreases until x = 0, increases
until x = 2, and decreases
thereafter. Thus, in interval
notation f is decreasing on
(, 0]  [2, ).
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Slide 4- 7
Extrema of Nonlinear Functions
Graphs of polynomial
functions often have
“hills” or “valleys”.
The “highest hill” on the graph is located at
(–2, 12.7). This is the absolute maximum of g.
There is a smaller peak located at the point
(3, 2.25). This is called the local maximum.
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Slide 4- 8
Extrema of Nonlinear Functions continued
Maximum and minimum values that are either absolute
or local are called extrema.
• A function may have several local extrema, but at
most one absolute maximum and one absolute
minimum.
• It is possible for a function to
assume an absolute
extremum at two values of x.
• The absolute maximum is 11.
It is a local maximum as well,
because near x = –2 and x = 2
it is the largest y-value.
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Slide 4- 9
Absolute and Local Extrema
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Slide 4- 10
Example
The monthly average ocean temperature in degrees
Fahrenheit at Bermuda can be modeled by
f ( x)  0.0215x4  0.648x3  6.03x2 17.1x  76.4,
where x = 1 corresponds to January and x = 12 to
December. The domain of f is D = {x|1  x  12 }.
(Source: J. Williams, The Weather Almanac 1995.)
a) Graph f in [1, 12, 1] by [50, 90, 10].
b) Estimate the absolute extrema. Interpret the results.
Solution
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Slide 4- 11
Solution continued
b) Many graphing calculators have the capability to
find maximum and minimum y-values.
•
•
An absolute minimum of about 61.5 corresponds
to the point (2.01, 61.5). This means the monthly
average ocean temperature is coldest during
February, when it reaches 61.5 F .
An absolute maximum of about 82 corresponds
to the point (7.61, 82.0), meaning that the
warmest average ocean temperature occurs in
August when it reaches a maximum of 82 F.
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Slide 4- 12
Symmetry
If a graph was folded along the y-axis, and the right and
left sides would match, the graph would be symmetric
with respect to the y-axis. A function whose graph
satisfies this characteristic is called an even function.
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Slide 4- 13
Symmetry continued
Another type of of symmetry occurs in respect to the
origin. If the graph could rotate, the original graph would
reappear after half a turn. This represents an odd
function.
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Slide 4- 14
Example
Identify whether the function is even or odd.
f  x   6x3  9x
Solution
Since f is a polynomial containing only odd
powers of x, it is an odd function. This also can
be shown symbolically as follows.
f x  6x  9x
3
 6 x 3  9 x
   6 x3  9 x 
  f  x
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Slide 4- 15
4.2
Polynomial Functions
and Models
♦ Understand the graphs of polynomial functions.
♦ Evaluate and graph piecewise-defined functions
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Graphs of Polynomial Functions
A polynomial function f of degree n can be
expressed as
f(x) = anxn + … + a2x2 + a1x + a0, where each
coefficient ak is a real number, an  0, and n is
a nonnegative integer.
A turning point occurs whenever the graph of a
polynomial function changes from increasing to
decreasing or from decreasing to increasing.
Turning points are associated with “hills” or
“valleys” on a graph.
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Slide 4- 17
Constant Polynomial Function
Has no x-intercepts or turning points
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Slide 4- 18
Linear Polynomial Function
Degree 1 and one x-intercept and no turning
points.
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Slide 4- 19
Quadratic Polynomial Functions
Degree 2, parabola that opens up or down. Can
have zero, one or two x-intercepts. Has exactly
one turning point, which is also the vertex.
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Slide 4- 20
Cubic Polynomial Functions
Degree 3, can have zero or two turning points.
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Slide 4- 21
Quartic Polynomial Functions
Degree 4, can have up to four x-intercepts and
three turning points.
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Slide 4- 22
Quintic Polynomial Functions
Degree 5, may have up to five x-intercepts and
four turning points.
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Slide 4- 23
Degree, x-intercepts, and turning points
The graph of a polynomial function of degree n  1 has
at most n x-intercepts and at most n  1 turning points.
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Slide 4- 24
Example
Use the graph of the polynomial
function shown.
a) How many turning points and
x-intercepts are there?
b) Is the leading coefficient a
positive or negative? Is the
degree odd or even?
c) Determine the minimum
degree of f.
Solution
a) There are three turning points corresponding to the
one “hill” and two “valleys”. There appear to be 4
x-intercepts.
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Slide 4- 25
Solution continued
b) Is the leading coefficient a
positive or negative? Is the
degree odd or even?
The left side and the right side
rise. Therefore, a > 0 and the
polynomial function has even
degree.
c) Determine the minimum degree of f.
The graph has three turning points. A polynomial of
degree n can have at most n  1 turning points.
Therefore, f must be at least degree 4.
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Slide 4- 26
Example
Graph f(x) = 2x3  5x2  5x + 7, and then complete the
following.
a) Identify the x-intercepts.
b) Approximate the coordinates of any turning
points to the nearest hundredth.
c) Use the turning points to approximate any local
extrema.
Solution
a) The graph appears to
intersect the x-axis at the
points (1.3, 0), (0.89, 0)
and (2.9, 0)
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Slide 4- 27
Solution continued
b) There are two turning points. From the
graphs their coordinates are approximately
(0.40, 8.1) and (2.07, 7.04)
c) There is a local maximum of about 8.07
and a local minimum of about 7.04.
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Slide 4- 28
Example
Let f(x) = 3x4 + 5x3  2x2.
a) Give the degree and leading coefficient.
b) State the end behavior of the graph of f.
Solution
a) The term with the highest degree is 3x4 so
the degree is 4 and the leading coefficient is 3.
b) The degree is even and the
leading coefficient is positive.
Therefore the graph of f
rises to the left and right.
More formally,
f ( x)   as x  
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Slide 4- 29
Piecewise-Defined Polynomial Functions
Example Evaluate f(x) at 6, 0, and 4.
5 x
 3
f ( x)  x  1
3  x 2

if x  5
if  4  x  2
if x  2
Solution
To evaluate f(6) we use the formula 5x
because 6 is < 5.
f(6) = 5(6) = 30
Similarly, f(0) = x3 + 1 = (0)3 + 1 = 1
f(4) = 3  x2 = 3  (4)2 = 13
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Slide 4- 30
Example
Complete the following.
a) Sketch the graph of f.
b) Determine if f is continuous on its domain.
c) Evaluate f(1).
4

f ( x) 4  x 2
2 x  6

if  4  x  0
if 0  x  2
if 2  x  4
Solution
a) Graph as shown to the right.
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Slide 4- 31
Solution continued
b) The domain is not continuous since there is a
break in the graph.
c) f(1) = 4  x2 = 4 – (1)2 = 3
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Slide 4- 32
4.3
Real Zeros of
Polynomial Functions
♦ Divide Polynomials
♦ Understand the division algorithm, remainder theorem,
and factor theorem
♦ Factor higher degree polynomials
♦ Analyze polynomials with multiple zeros
♦ Find rational zeros
♦ Solve polynomial equations
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Example
Divide x3 + 2x2  5x  6 by x  3. Check the result.
Solution
2
x  5 x  10
3
2
x  3 x  2 x  5x  6
x3  3x 2
5x  5x
2
5 x 2  15 x
10 x  6
10 x  30
The quotient is x2 + 5x + 10
with a remainder of 24.
24
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Slide 4- 34
Solution continued
Check
( x  3)( x 2  5 x  10)  24  x( x 2  5 x  10)  3( x 2  5 x  10)  24
 x3  5 x 2  10 x  3x 2  15 x  30  24
 x3  2 x 2  5 x  6
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Slide 4- 35
Synthetic Division
A short cut called synthetic division can be used to divide
x – k into a polynomial.
Steps
1. Write k to the left and the coefficients of f(x) to the right
in the top row. If any power does not appear in f(x),
include a 0 for that term.
2. Copy the leading coefficient of f(x) into the third row and
multiply it by k. Write the result below the next
coefficient of f(x) in the second row. Add the numbers in
the second column and place the result in the third row.
Repeat the process.
3. The last number in the third row is the remainder. If the
remainder is 0, then the binomial x – k is a factor of f(x).
The other numbers in the third row are the coefficients
of the quotient in descending powers.
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Slide 4- 36
Example
Use synthetic division to divide 2x3 + 7x2 – 5 by
x + 3.
Solution Let k = –3 and perform the following.
–3
7
0 –5
–6 –3
9
2
1 –3
4
2
The remainder is 4 and the quotient is
2x2 + x – 3. The result can be expressed as
4
2x  x  3 
x3
2
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Slide 4- 37
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Slide 4- 38
Example
Use the graph of f(x) = x3  x2 – 9x + 9 and the
factor theorem to list the factors of f(x).
Solution
The graph shows that the
zeros or x-intercepts of f
are 3, 1and 3.
Since f(3) = 0, the factor
theorem states that(x + 3)
is a factor, and f(1) = 0
implies that (x  1) is a factor and f(3) = 0
implies (x  3) is a factor. Thus the factors are
(x + 3)(x  1), and (x  3).
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Slide 4- 39
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Slide 4- 40
Example
Write the complete factorization for the
polynomial 6x3 + 19x2 + 2x – 3 with given zeros
–3, –1/2 and 1/3.
Solution
Leading coefficient is 6
Zeros are –3, –1/2 and 1/3
The complete factorization:
1 
1

f ( x)  6( x  3)  x   x  
2 
3

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Slide 4- 41
Example
The polynomial f(x) = 2x3  3x2  17x + 30 has a
zero of 2. Express f(x) in complete factored
form.
Solution
If 2 is a zero, by the factor theorem x  2 is a
factor. Use synthetic division.
2
2
2
– 3 –17 30
4
2 –30
1 –15
0
( x  2)(2x  x 15)  ( x  2)(2x  5)( x  3)
2
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Slide 4- 42
Rational Zeros
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Slide 4- 43
Example
Find all rational zeros of
f(x) = 6x4 + 7x3  12x2  3x + 2. Write in complete
factored form.
Solution
p : 1
q : 1
2
2
3
6
Any rational zero must occur in the list
1
 ,
1
2
 ,
1
1
 ,
2
1
1
 ,  ,
3
6
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2

3
Slide 4- 44
Solution continued
Evaluate f(x) at each value in the list.
x
f(x)
x
f(x)
x
f(x)
1
1
2
2
0
8
100
0
½
½
1/3
1/3
5/4
0
0
1.48
1/6
1/6
2/3
2/3
1.20
2.14
2.07
2.22
From the table there are four rational zeros of 1,
2, 1/2, and 1/3. The complete factored form
is:
1 
1

6( x  1)( x  2)  x   x  
2 
3

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Slide 4- 45
Example
Find all real solutions to the equation
4x4 – 17x2 – 50 = 0.
Solution
4
2
4
x

17
x
 50  0
The expression can be
2
2
(4
x

25)(
x
 2)  0
factored similar to a
2
2
4
x

25

0
or
x
20
quadratic equation.
5
The only solutions are  2
since the equation x2 = –2
has no real solutions.
4 x 2  25
25
2
x 
4
5
x
2
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x 2  2
x 2  2
x 2  2
Slide 4- 46
Example
Solve the equation x3 – 2.1x2 – 7.1x + 0.9 = 0
graphically. Round any solutions to the nearest
hundredth.
Solution
Since there are three x-intercepts the equation has
three real solutions.
x  .012, 1.89, and 3.87
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Slide 4- 47
4.4
The Fundamental
Theorem of Algebra
♦ Perform arithmetic operations on complex numbers
♦ Solve quadratic equations having complex solutions
♦ Apply the fundamental theorem of algebra
♦ Factor polynomials having complex zeros
♦ Solve polynomial equations having complex solutions
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Complex Numbers
A complex number can be written in standard
form as a + bi, where a and b are real
numbers. The real part is a and the imaginary
part is b.
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Slide 4- 49
Example
Write each expression in standard form.
Support your results using a calculator.
a) (4 + 2i) + (6  3i) b) (9i)  (4  7i)
c) (2 + 5i)2
d) 16
3i
Solution
a) (4 + 2i) + (6  3i) = 4 + 6 + 2i  3i = 2  i
b) (9i)  (4  7i) = 4  9i + 7i = 4  2i
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Slide 4- 50
Solution continued
c) (2 + 5i)2 = (2 + 5i)(2 + 5i)
= 4 – 10i – 10i + 25i2
= 4  20i + 25(1)
= 21  20i
16
16 16


d)
3i 3i 3i
48  16i

9  i2
48  16i 24 8


 i
10
5 5
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Slide 4- 51
Quadratic Equations with Complex Solutions
We can use the quadratic formula to solve
quadratic equations if the discriminant is
negative.
There are no real solutions, and the graph does
not intersect the x-axis.
The solutions can be expressed as imaginary
numbers.
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Slide 4- 52
Example
Solve the quadratic equation 4x2 – 12x = –11.
Solution
Rewrite the equation: 4x2 – 12x + 11 = 0
b  b 2  4ac
a = 4, b = –12, c = 11
x
2a
12  (12) 2  4(4)(11)

2(4)
12  32

8
12  4i 2 3 i 2

 
8
2
2
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Slide 4- 53
Fundamental Theorem of Algebra
The polynomial f(x) of degree n  1 has at least one
complex zero.
Number of Zeros Theorem
A polynomial of degree n has at most n distinct zeros.
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Slide 4- 54
Example
Represent a polynomial of degree 4 with
leading coefficient 3 and zeros of 2, 4, i and i
in complete factored form and expanded form.
Solution
Let an = 3, c1 = 2, c2 = 4, c3 = i, and c4 = i.
f(x) = 3(x + 2)(x  4)(x  i)(x + i)
Expanded: 3(x + 2)(x  4)(x  i)(x + i)
= 3(x + 2)(x  4)(x2 + 1)
= 3(x + 2)(x3  4x2 + x  4)
= 3(x4  2x3  7x2  2x  8)
= 3x4  6x3 – 21x2 – 6x – 24
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Slide 4- 55
Conjugate Zeros Theorem
If a polynomial f(x) has only real coefficients and if
a + bi is a zero of f(x), then the conjugate a  bi is also a
zero of f(x).
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Slide 4- 56
Example
Find the zeros of f(x) = x4 + 5x2 + 4 given one
zero is i.
Solution
By the conjugate zeros theorem it follows that i
must be a zero of f(x).
(x + i) and (x  i) are factors
(x + i)(x  i) = x2 + 1, using long division we can
find another quadratic factor of f(x).
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Slide 4- 57
Solution continued
Long division
x2  4
x 2  0 x  1 x 4  0 x3  5 x 2  0 x  4
x 4  0 x3  x 2
4 x2  0 x  4
4 x2  0 x  4
0
The solution is
x4 + 5x2 + 4 = (x2 + 4)(x2 + 1)
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Slide 4- 58
Example
Solve x3 = 2x2  5x + 10.
Solution
Rewrite the equation: f(x) = 0, where
f(x) = x3  2x2 + 5x  10
We can use factoring by grouping or graphing
to find one real zero.
The graph shows a zero
at 2. So, x  2 is a factor.
Use synthetic division.
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Slide 4- 59
Solution continued
Synthetic division
2
1 –2
1
2
0
5 –10
0
5
10
0
x3  2x2 + 5x  10 = (x  2)(x2 + 5)
x20
or
x2
or
x2
or
x2  5  0
x  5
2
x  i 5
The solutions are 2 and x  i 5.
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Slide 4- 60
4.5
Rational Functions
and Models
♦ Identify a rational function and state its domain
♦ Find and interpret vertical asymptotes
♦ Find and interpret horizontal asymptotes
♦ Solve rational equations
♦ Solve applications involving rational equations
♦ Solve applications involving variation
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Example
1
Use the graph of f ( x)  2 to sketch the graph of
x
1
g ( x) 
 2. Include all asymptotes in your
( x  1)
2
graph. Write g(x) in terms of f(x).
Solution
g(x) is a translation of f(x)
left one unit and down 2 units.
The vertical asymptote is
x = 1
The horizontal asymptote is
y = 2
g(x) = f(x + 1)  2
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Slide 4- 64
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Slide 4- 65
Example
For each rational function, determine any
horizontal or vertical asymptotes.
2
2x  6
x 1
x
a) f ( x) 
b) f ( x)  2
c) f ( x)   4
4x  8
x 9
x2
Solution
a) Horizontal Asymptote: Degree of numerator
equals the degree of the
denominator.
y = a/b is asymptote,
so y = 2/4 = 1/2
Vertical Asymptote:
4x  8 = 0, x = 2
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Slide 4- 66
Solution continued b)
x 1
f ( x)  2
x 9
Horizontal Asymptote:
Degree: numerator < denominator
y=0
Vertical Asymptote:
x2  9 = 0
x =  3
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Slide 4- 67
Solution continued c)
x2  4
f ( x) 
x2
Horizontal Asymptote:
Degree: numerator > denominator
no horizontal asymptotes
Vertical Asymptote:
no vertical asymptotes
x2  4
f ( x) 
x2
( x  2)( x  2)

x2
 x2
x2
The graph is the line y = x + 2 with the point (2, 4) missing.
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Slide 4- 68
Slant or Oblique Asymptote
A third type of asymptote is neither horizontal or
vertical.
Occurs when the numerator
of a rational function
has a degree one more
than the degree of
the denominator.
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Slide 4- 69
Example
x2  1
.
Let f ( x) 
x2
a) Use a calculator to graph f.
b) Identify any asymptotes.
c) Sketch a graph of f that includes the
asymptotes.
Solution
a)
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Slide 4- 70
Solution continued
b) Asymptotes: The function is undefined
when x  2 = 0 or when x = 2.
Vertical asymptote at x = 2
Oblique asymptote at y = x + 2
c)
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Slide 4- 71
Example
2x
 4.
Solve
x2
Solution
Symbolic
Graphical
Numerical
2x
4
x2
2 x  4( x  2)
2x  4x  8
2 x   8
x4
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Slide 4- 72
Example
4
2
3

.
Solve 2 
x 1 x 1 x 1
Solution
Multiply by the LCD to clear the fractions.
4
2
3


2
x 1 x 1 x 1
4( x  1)( x  1) 2( x  1)( x  1) 3( x  1)( x  1)


2
x 1
x 1
x 1
4  2( x  1)  3( x  1)
4  2 x  2  3x  3
4  5x 1
When 1 is substituted for x, two expressions
1 x
in the given equation are undefined.
There are no solutions.
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Slide 4- 73
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Slide 4- 74
Example
At a distance of 3 meters, a 100-watt bulb
produces an intensity of 0.88 watt per square
meter.
a) Find the constant of proportionality k.
b) Determine the intensity at a distance of 2.5
meters.
Solution
a) Substitute d = 3 and I = 0.88 into the
equation and solve for k.
k
I
d2
0.88 
k
or k  7.92
2
3
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Slide 4- 75
Solution continued
b) Let
I
7.92
d2
and d = 2.5.
7.92
I 2
d
7.92
I
 1.27
2
2.5
The intensity at 2.5 meters is 1.27 watts per
square meter.
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Slide 4- 76
4.6
Polynomial and
Rational Inequalities
♦ Solve polynomial inequalities
♦ Solve rational inequalities
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction
• An inequality says that one expression is
greater than, greater than or equal to,
less than, or less than or equal to,
another expression.
• Solving Inequalities
• Boundary numbers (x-values) are
found where the inequality holds.
• A graph or a table of test values can
be used to determine the intervals
where the inequality holds.
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Slide 4- 78
Polynomial Inequalities
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Slide 4- 79
Example
Solve x3  7 x 2  10 x symbolically and graphically.
Solution Symbolically
Step 1: Write the inequality as x3  7 x2  10 x  0.
Step 2: Replace the inequality symbol with an
equal sign and solve.
x3  7 x 2  10 x  0
x  x 2  7 x  10   0
x  x  5 x  2   0
The boundary numbers
are –5, –2, and 0.
x  0 or x  5 or x  2
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Slide 4- 80
Solution continued
Step 3: The boundary numbers separate the
number line into four disjoint intervals:
 , 5 ,  5, 2 ,  2,0 , and 0, 
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
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Slide 4- 81
Solution continued
Step 4: Complete a table of test values.
Interval Test Value x x3 + 7x2 + 10x Positive/Negative
 , 5
 5, 2
–6
–24
Negative
–4
8
Positive
 2,0
 0,
–1
–4
Negative
1
18
Positive
The solution set is [5, 2]
0, .
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Slide 4- 82
Solution continued
Graphically
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Slide 4- 83
Rational Inequalities
• Inequalities involving rational expressions are
called rational inequalities.
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Slide 4- 84
Example
Health Care: There is a formula for determining
the amount of a child's particular medication
dosage based on an adults dosage. Solve the
inequality to determine the age of the child to
receive the dose of 8 mg when an adult
receives 24 mg. (Source: Olsen, Medical Dosage Calculations, 6 ed.)
th
a
8
24
a  12
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Slide 4- 85
Example continued
Solution
Solving this graphically, we find this dose is
acceptable when the child is at least 6 years
of age.
a
8
24
a  12
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Slide 4- 86
Example
5
 1.
Solve
x4
Solution
p  x
Step 1: Rewrite the inequality in the form q  x   0.
5
1
x4
5
1  0
x4
5   x  4
0
x4
1 x
0
x4
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Slide 4- 87
Solution continued
Step 2: Find the zeros of the numerator and
the denominator.
Numerator
1–x=0
x=1
Denominator
x+4=0
x = –4
Step 3: The boundary numbers are – 4 and 1,
which separate the number line into three
disjoint intervals:  , 4 ,  4,1 and 1, .
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Slide 4- 88
Solution continued
Step 4: Use a table to solve the inequality.
Interval Test Value x
 , 4
 4,1
1,
(1–x)/(x + 4)
Positive/Negative
–5
–6
Negative
–2
3/2
Positive
2
–1/6
Negative
The interval notation is (–4, 1].
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Slide 4- 89
Rational Inequality
• Caution: When solving a rational inequality, it
is essential not to multiply or divide each side
of the inequality by the LCD if the LCD
contains a variable. This techniques often
leads to an incorrect solution set.
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Slide 4- 90
4.7
Power Functions and
Radical Equations
♦ Learn properties of rational exponents
♦ Learn radical notation
♦ Understand properties and graphs of power functions
♦ Use power functions to model data
♦ Solve equations involving rational exponents
♦ Solve equations involving radical expressions
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Rational Exponents and Radical Notation
• Expressions with rational exponents can be
simplified with the following properties.
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Slide 4- 92
Example
Simplify each expression by hand.
a) 82/3
b) (–32)–4/5
Solutions
•
•
2/ 3
8

 32
 8
3
4/ 5


2
5
2 4
2
32

4

1
 2
4
1

16
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Slide 4- 93
Example
Use positive rational exponents to write each
expression.
a) 5 4
b)
3
6
x
x
Solutions
a) 5 x 4 
x 
b)
x  x
3
x
 x
6

4 1/ 5
x
1/ 3
1/ 2 1/ 2
4/5

1/ 6 1/ 2
x
x

 x
1/ 31/ 6

1/ 2
 x1/ 4
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Slide 4- 94
Power Functions and Models
• Power functions typically have rational
exponents.
• A special type of power function is a root
function.
• Examples of power functions include:
f1(x) = x2, f2(x) = x3/4 , f3(x) = x0.4, and f4(x) =
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3
x2
Slide 4- 95
Solution continued
• Often, the domain of a power function f is restricted to
nonnegative numbers.
• Suppose the rational number p/q is written in lowest
terms. The the domain of f(x) = xp/q is all real numbers
whenever q is odd and all nonnegative numbers
whenever q is even.
• The following graphs show 3 common power functions.
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Slide 4- 96
Example
•
Modeling Wing Size of a Bird: Heavier
birds have larger wings with more surface
areas than do lighter birds. For some
species the relationship can be modeled by
S(w) = 0.2w2/3, where w is the weight of the
bird in kilograms and S is surface area of the
wings in square meters. (Source: C. Pennycuick,
Newton Rules Biology.)
a) Approximate S(0.75) and interpret the result.
b) What weight corresponds to a surface area
of 0.45 square meter?
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Slide 4- 97
Example continued
Solution
a) S(0.75) = 0.2(0.75)2/3  0.165. The wings of a
bird that weighs about 0.75 kilogram have
the surface area of about 0.165 square
meter.
b) To answer this, we must solve the equation
0.2w2/3 = 0.45.
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Slide 4- 98
Solution continued
0.2 w2 / 3  0.45
0.45
2/3
w 
0.2
w 
2/3 3
 0.45 


0.2


 0.45 
w 

0.2


3
3
2
 0.45 
w 

0.2


w  3.4
3
Since w must be positive,
the wings of a 3.4 kilogram
bird must have a surface
area of about 0.45 square
meter.
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Slide 4- 99
Equations Involving Rational Exponents
Example Solve 4x3/2 – 6 = 6. Approximate the
answer to the nearest hundredth, and give
graphical support.
Solutions
Symbolic Solution
Graphical Solution
4x3/2 – 6 = 6
4x3/2 = 12
(x3/2)2 = 32
x3 = 9
x = 91/3
x = 2.08
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Slide 4- 100
Equations Having Negative Exponents
Example Solve 6 x 2  x 1  2.
Solution
6 x 2  x 1  2
6u 2  u  2  0
 3u  2  2u  1  0
2
1
u   or u 
3
2
3
x   or x  2
2
1
1
Since u  , then x  .
x
u
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Slide 4- 101
Equations Involving Radicals
When solving equations that contain square
roots, it is common to square each side of an
equation.
Example Solve 3x  2  x  2.
Solution
3x  2  x  2

3x  2

2
  x  2
2
3x  2  x 2  4 x  4
x  7x  6  0
2
 x  1 x  6   0
x  1 or x  6
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Slide 4- 102
Solution continued
Check
3 1  2  1  2
1  1
36  2  6  2
44
Substituting these values in the original
equation shows that the value of 1 is an
extraneous solution because it does not
satisfy the given equation.
• Therefore, the only solution is 6.
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Slide 4- 103
Example
Some equations may contain a cube root.
Solve 3 4x2  4x 1  3 x.
Solution
3
2
3
4x  4x 1  x

3
  x
3
4x  4x 1 
2
3
3
4 x2  5x  1  0
 4 x  1 x  1  0
1
x
or x  1
4
1
Both solutions check, so the solution set is  ,
4
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
1 .

Slide 4- 104