Transcript florist 52

Chapter 0
Functions
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 78
Chapter Outline

Functions and Their Graphs

Some Important Functions

The Algebra of Functions

Zeros of Functions – The Quadratic Formula and Factoring

Exponents and Power Functions

Functions and Graphs in Applications
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 2 of 78
§ 0.1
Functions and Their Graphs
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 3 of 78
Section Outline

Rational and Irrational Numbers

The Number Line

Open and Closed Intervals

Applications of Functions

Domain of a Function

Graphs of Functions

The Vertical Line Test

Graphing Calculators

Graphs of Equations
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 4 of 78
Rational & Irrational Numbers
Definition
Rational Number: A number
that may be written as a finite
or infinite repeating decimal,
in other words, a number that
can be written in the form
m/n such that m, n are
integers
Irrational Number: A
number that has an infinite
decimal representation whose
digits form no repeating
pattern
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Example

2
 0.285714
7
3  1.73205
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 5 of 78
The Number Line
The Number Line
A geometric representation of the real numbers is shown
below.

-6
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-5
-4
-3
-2
-1
2
7
0
3
1
2
3
4
5
6
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 6 of 78
Open & Closed Intervals
Definition
Open Interval: The set of
numbers that lie between
two given endpoints, not
including the endpoints
themselves
Closed Interval: The set of
numbers that lie between
two given endpoints,
including the endpoints
themselves
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Example
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
2
3
4
5
6
4, 
x4
-6
-5
-4
-3
-2
-1
0
1
[-1, 4]
1  x  4
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 7 of 78
Functions in Application
EXAMPLE
(Response to a Muscle) When a solution of acetylcholine is introduced into the heart muscle of
a frog, it diminishes the force with which the muscle contracts. The data from experiments of
the biologist A. J. Clark are closely approximated by a function of the form
Rx  
100 x
b x
where x is the concentration of acetylcholine (in appropriate units), b is a positive constant that
depends on the particular frog, and R(x) is the response of the muscle to the acetylcholine,
expressed as a percentage of the maximum possible effect of the drug.
(a) Suppose that b = 20. Find the response of the muscle when x = 60.
(b) Determine the value of b if R(50) = 60 – that is, if a concentration of x = 50 units produces a
60% response.
SOLUTION
(a)
Rx  
100 x
b x
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This is the given function.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 8 of 78
Functions in Application
CONTINUED
R60 
100  60
20  60
Replace b with 20 and x with 60.
R60  
6000
80
Simplify the numerator and
denominator.
R60  75
Divide.
Therefore, when b = 20 and x = 60, R (x) = 75%.
(b)
Rx  
100 x
b x
This is the given function.
R50 
100  50
b  50
Replace x with 50.
60 
100  50
b  50
Replace R(50) with 60.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 9 of 78
Functions in Application
CONTINUED
60 
b  50 60 
5000
b  50
Simplify the numerator.
5000
 b  50
b  50
Multiply both sides by b + 50 and cancel.
60b  3000  5000
Distribute on the left side.
60b  2000
Subtract 3000 from both sides.
b  33.3
Divide both sides by 60.
Therefore, when R (50) = 60, b = 33.3.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 10 of 78
Functions
EXAMPLE
If f x   x 2  4 x  3 , find f (a - 2).
SOLUTION
f x   x 2  4 x  3
This is the given function.
f a  2  a  2  4a  2  3
2


Replace each occurrence of x with a – 2.
f a  2  a 2  4a  4  4a  2  3
Evaluate (a – 2)2 = a2 – 4a + 4.
f a  2  a 2  4a  4  4a  8  3
Remove parentheses and distribute.
f a  2  a 2  1
Combine like terms.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 11 of 78
Domain
Definition
Domain of a Function: The
set of acceptable values for
the variable x.
Example
The domain of the function
f x  
is
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1
3 x
3 x  0
3 x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 12 of 78
Graphs of Functions
Definition
Example
Graph of a Function: The set
of all points (x, f (x)) where x is
the domain of f (x). Generally,
this forms a curve in the xyplane.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 13 of 78
The Vertical Line Test
Definition
Example
Vertical Line Test: A curve in
the xy-plane is the graph of a
function if and only if each
vertical line cuts or touches the
curve at no more than one
point.
Although the red line intersects
the graph no more than once
(not at all in this case), there
does exist a line (the yellow
line) that intersects the graph
more than once. Therefore,
this is not the graph of a
function.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 14 of 78
Graphing Calculators
Graphing Using a Graphing Calculator
Step
Display
1) Enter the expression
for the function.
2) Enter the specifications
for the viewing window.
3) Display the graph.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 15 of 78
Graphs of Equations
EXAMPLE


1
2
Is the point (3, 12) on the graph of the function f x    x  x  2 ?
SOLUTION
1

f  x    x   x  2 
2

This is the given function.
1

f 3   3  3  2 
2

Replace x with 3.
1

12   3  3  2
2

Replace f (3) with 12.
12  2.55
Simplify.
12  12.5
false
Multiply.
Since replacing x with 3 and f (x) with 12 did not yield a true statement in the original function, we
conclude that the point (3, 12) is not on the graph of the function.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 16 of 78
§ 0.2
Some Important Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 17 of 78
Section Outline

Linear Equations

Applications of Linear Functions

Piece-Wise Functions

Quadratic Functions

Polynomial Functions

Rational Functions

Power Functions

Absolute Value Function
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 18 of 78
Linear Equations
Equation
Example
y = mx + b
(This is a linear function)
x=a
(This is not the graph of a
function)
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 19 of 78
Linear Equations
CONTINUED
Equation
Example
y=b
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 20 of 78
Applications of Linear Functions
EXAMPLE
(Enzyme Kinetics) In biochemistry, such as in the study of enzyme kinetics, one encounters a linear
function of the form f x  K / V x  1/ V , where K and V are constants.
(a) If f (x) = 0.2x + 50, find K and V so that f (x) may be written in the form, f x  K / V x  1/ V.
(b) Find the x-intercept and y-intercept of the line f x  K / V x  1/ V in terms of K and V.
SOLUTION
(a) Since the number 50 in the equation f (x) = 0.2x + 50 is in place of the term 1/V (from the
original function), we know the following.
50 = 1/V
50V = 1
V = 0.02
Explained above.
Multiply both sides by V.
Divide both sides by 50.
Now that we know what V is, we can determine K. Since the number 0.2 in the equation
f (x) = 0.2x + 50 is in place of K/V (from the original function), we know the following.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 21 of 78
Applications of Linear Functions
CONTINUED
0.2 = K/V
0.2V = K
0.2(0.02) = K
0.004 = K
Explained above.
Multiply both sides by V.
Replace V with 0.02.
Multiply.
Therefore, in the equation f (x) = 0.2x + 50, K = 0.004 and V = 0.02.
(b) To find the x-intercept of the original function, replace f (x) with 0.
f x  K / V x  1/ V
0  K / V x  1/ V
1/ V  K / V x
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This is the original function.
Replace f (x) with 0.
Solve for x by first subtracting 1/V from
both sides.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 22 of 78
Applications of Linear Functions
CONTINUED
 1/ V V / K   x
1 / K  x
Multiply both sides by V/K.
Simplify.
Therefore, the x-intercept is -1/K. To find the y-intercept of the original function, we
recognize that this equation is in the form y = mx + b. Therefore we know that 1/V is the
y-intercept.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 23 of 78
Piece-Wise Functions
EXAMPLE
1  x for x  3
.
2
for
x

3

Sketch the graph of the following function f x   
SOLUTION
We graph the function f (x) = 1 + x only for those values of x that are less than or equal to 3.
6
4
2
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-2
-4
-6
Notice that for all values of x greater than 3, there is no line.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 24 of 78
Piece-Wise Functions
CONTINUED
Now we graph the function f (x) = 4 only for those values of x that are greater than 3.
6
5
4
3
2
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Notice that for all values of x less than or equal to 3, there is no line.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 25 of 78
Piece-Wise Functions
CONTINUED
Now we graph both functions on the same set of axes.
-6
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-5
-4
-3
-2
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
1
2
3
4
5
6
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 26 of 78
Quadratic Functions
Definition
Example
Quadratic Function:
A function of the
form
f x   ax 2  bx  c
where a, b, and c are
constants and a  0.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 27 of 78
Polynomial Functions
Definition
Example
Polynomial Function: A
function of the form
f x   an x n  an 1 x n 1    a0
f x   17 x3  x 2  5
where n is a nonnegative
integer and a0, a1, ...an are
given numbers.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 28 of 78
Rational Functions
Definition
Rational Function: A
function expressed as the
quotient of two
polynomials.
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Example
3x  x 4
g x   2
5x  x  1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 29 of 78
Power Functions
Definition
Example
Power Function: A
function of the form
f x   x 5.2
f x   x r .
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 30 of 78
Absolute Value Function
Definition
Example
Absolute Value Function:
The function defined for
all numbers x by
f x   x
f x   x ,
such that |x| is understood
to be x if x is positive and
–x if x is negative
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f 1 2  1 2  1 2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 31 of 78
§ 0.3
The Algebra of Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 32 of 78
Section Outline

Adding Functions

Subtracting Functions

Multiplying Functions

Dividing Functions

Composition of Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 33 of 78
Adding Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express f (x) + g (x) as a rational function.
x3
x2
SOLUTION
f (x) + g (x) =

2
1

x 3 x  2
Replace f (x) and g (x) with the given
functions.

x2 2
1 x3



x  2 x 3 x  2 x 3
Multiply to get common denominators.

2x  4
x 3

x  2x  3 x  2x  3
Evaluate.

2x  4  x  3
x  2x  3
Add.

3x  1
x  2x  3
Simplify the numerator.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 34 of 78
Adding Functions
CONTINUED

3x  1
x 2  3x  2 x  6
Evaluate the denominator.

3x  1
x  x6
Simplify the denominator.
2
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 35 of 78
Subtracting Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express f (x) - g (x) as a rational function.
x3
x2
SOLUTION
f (x) - g (x) =

2
1

x 3 x  2
Replace f (x) and g (x) with the given
functions.

x2 2
1 x 3



x  2 x 3 x  2 x 3
Multiply to get common denominators.

2x  4
x 3

x  2x  3 x  2x  3
Evaluate.

2 x  4  x  3
x  2x  3
Subtract.

x7
x  2x  3
Simplify the numerator.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 36 of 78
Subtracting Functions
CONTINUED

x7
x 2  3x  2 x  6
Evaluate the denominator.

x7
x  x6
Simplify the denominator.
2
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 37 of 78
Multiplying Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express f (x)g (x) as a rational function.
x3
x2
SOLUTION
f (x)g (x) =

2
1

x 3 x  2
Replace f (x) and g (x) with the given
functions.

2 1
x  3x  2
Multiply the numerators and
denominators.

2
x  x6
Evaluate.
2
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 38 of 78
Dividing Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express [f (x)]/[g (x)] as a rational function.
x3
x2
SOLUTION
f (x)/g (x) =
2
 x 3
1
x2
Replace f (x) and g (x) with the given
functions.

2 x2

x 3 1
Rewrite as a product (multiply by
reciprocal of denominator).

2 x  2
x  31
Multiply the numerators and
denominators.

2x  4
x3
Evaluate.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 39 of 78
Composition of Functions
EXAMPLE
(Conversion Scales) Table 1 shows a conversion table for men’s hat sizes for three countries. The
function g x  8x  1 converts from British sizes to French sizes, and the function f  x   1 x
converts from French sizes to U.S. sizes. Determine the function h (x) = f (g (x)) and give 8
its interpretation.
SOLUTION
h (x) = f (g (x))
This is what we will determine.
1
  g x 
8
In the function f, replace each
occurrence of x with g (x).
1
  8 x  1
8
Replace g (x) with 8x + 1.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 40 of 78
Composition of Functions
CONTINUED
1
1
  8 x  1
8
8
Distribute.
1
8
Multiply.
 x
Therefore, h (x) = f (g (x)) = x + 1/8. Now to determine what this function h (x) means, we must
recognize that if we plug a number into the function, we may first evaluate that number plugged
into the function g (x). Upon evaluating this, we move on and evaluate that result in the function
f (x). This is illustrated as follows.
g (x)
British
f (x)
French
French
U.S.
h (x)
Therefore, the function h (x) converts a men’s British hat size to a men’s U.S. hat size.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 41 of 78
§ 0.4
Zeros of Functions – The Quadratic Formula
and Factoring
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 42 of 78
Section Outline

Zeros of Functions

Quadratic Formula

Graphs of Intersecting Lines

Factoring
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 43 of 78
Zeros of Functions
Definition
Example
Zero of a Function: For
a function f (x), all values
of x such that f (x) = 0.
f x   x 2  1
0  x2 1
x  1
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 44 of 78
Quadratic Formula
Definition
Quadratic Formula: A
formula for solving any
quadratic equation of the
form ax 2  bx  c  0 .
The solution is:
 b  b  4ac
x
.
2a
2
There is no solution if
b 2  4ac  0.
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Example
x 2  3x  2  0
a  1; b  3; c  2
x
x
 3 
32  41 2
21
 3  17
2
These are the
solutions/zeros of the
quadratic function
f x   x 2  3x  2.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 45 of 78
Graphs of Intersecting Functions
EXAMPLE
Find the points of intersection of the pair of curves.
y  x 2  10 x  9;
y  x9
SOLUTION
The graphs of the two equations can be seen to intersect in the following graph. We can use this
graph to help us to know whether our final answer is correct.
100
80
60
40
20
0
-5
0
5
10
15
-20
-40
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 46 of 78
Graphs of Intersecting Functions
CONTINUED
To determine the intersection points, set the equations equal to each other, since they both equal the
same thing: y.
x 2  10 x  9  x  9
Now we solve the equation for x using the quadratic formula.
x 2  10 x  9  x  9
This is the equation to solve.
x 2  11x  9  9
Subtract x from both sides.
x 2  11x  18  0
Add 9 to both sides.
We now recognize that, for the quadratic formula, a = 1, b = -11, and c =18.
x
x
  11 
 112  4118
21
11  121  72
2
© 2010 Pearson Education Inc.
Use the quadratic formula.
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 47 of 78
Graphs of Intersecting Functions
CONTINUED
x
11 49
2
Simplify.
x
11 7
2
Simplify.
x
11  7 11  7
,
2
2
Rewrite.
x  9, 2
Simplify.
We now find the corresponding y-coordinates for x = 9 and x = 2. We can use either of the original
equations. Let’s use y = x – 9.
x9
x2
y  x9
y  x9
y  99
y  29
y0
y  7
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 48 of 78
Graphs of Intersecting Functions
CONTINUED
Therefore the solutions are (9, 0) and (2, -7). This seems consistent with the two intersection
points on the graph. A zoomed in version of the graph follows.
10
5
0
0
2
4
6
8
10
-5
-10
-15
-20
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 49 of 78
Factoring
EXAMPLE
Factor the following quadratic polynomial.
6 x  2 x3
SOLUTION
6 x  2 x3

2x 3  x2
This is the given polynomial.

Factor 2x out of each term.
2
2 x 3  x 2 


Rewrite 3 as
2
3 .
Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b).

2x 3  x

3x
© 2010 Pearson Education Inc.

2
Rewrite 3  x 2 as

3x


3x .
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 50 of 78
Factoring
EXAMPLE
Solve the equation for x.
1
5 6

x x2
SOLUTION
1
5 6

x x2
5 6 
x 2 1    2  x 2
x x 
x2 
5 2 6 2
x  2 x
x
x
x 2  5x  6
x 2  5x  6  0
x 1x  6  0
© 2010 Pearson Education Inc.
This is the given equation.
Multiply everything by the LCD: x2.
Distribute.
Multiply.
Subtract 5x + 6 from both sides.
Factor.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 51 of 78
Factoring
CONTINUED
x 1  0
x  1
x6  0
x6
Set each factor equal to zero.
Solve.
Since no denominator from the original equation is zero when x = -1 or when x = 6, these are our
solutions.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 52 of 78
§ 0.5
Exponents and Power Functions
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 53 of 78
Section Outline

Exponent Rules

Applications of Exponents

Compound Interest
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 54 of 78
Exponents
Definition
Example
b n  b
 b



b
53  5  5  5
n times
1
n
b  b
© 2010 Pearson Education Inc.
n
1
3
5 3 5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 55 of 78
Exponents
Definition
m
n
b  b 
b

m
n

m
n
1
b
m
n
© 2010 Pearson Education Inc.

 b
1
n
b
Example
m
3
4
5  5 
m
n

1
 b
n
m
5

3
4

3
4
1
5
3
4

 5
1
4
3
5
4

3
1
 5
4
3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 56 of 78
Exponents
Definition
b b  b
r
b
© 2010 Pearson Education Inc.
s
r
1
 r
b
rs
Example
1
3
2
3
6 6  6
4

1
2

1 2

3 3
1
4
1
2
3
3
 6  61  6

1
1

4 2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 57 of 78
Exponents
Definition
Example
7
br
r s

b
bs
b 
r s
© 2010 Pearson Education Inc.
 b rs
7

9


4
5
4
3
1
3
7
4 1

3 3
3
3
 7  71  7
5
8
45
4
1


  95 8  98  92  9  3


Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 58 of 78
Exponents
Definition
abr  a r  b r
r
ar
a
   r
b b
© 2010 Pearson Education Inc.
Example
125  271/ 3  1251/ 3  271/ 3
 3 125  3 27  5  3  15
4
10 4  10 
4




2
 16


4
5
5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 59 of 78
Applications of Exponents
EXAMPLE
Use the laws of exponents to simplify the algebraic expression.
 27 x 
5 2/3
3
x
SOLUTION
 27 x 
5 2/3
3
x
 27 2 / 3  x 5 2 / 3
3
1/ 3
x
 27 2 / 3  x10 / 3
1/ 3
3
x
abr  a r  b r
x
 272 / 3  x5 2 / 3

This is the given expression.

2
 27  x10 / 3
x1/ 3
© 2010 Pearson Education Inc.
1
n
b n b
b 
r s
m
n
 b rs
b  n bm 
 b
n
m
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 60 of 78
Applications of Exponents
CONTINUED
 32  x10 / 3
x1 / 3
9 x10 / 3
x1/ 3
3
 27  3
 32  9
9x
br
 br s
s
b
9x 9 / 3
Subtract.
9x 3
Divide.
10 / 31/ 3
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 61 of 78
Compound Interest - Annual
Definition
Example
Compound Interest Formula:
A  P1  i 
n
If $700 is invested, compounded
annually at 8% for 8 years, this will
grow to:
A = the compound amount
8
(how much money you end
A  7001  0.08
up with)
8
A  7001.08
P = the principal amount
A  7001.851
invested
A  1,295.651
i = the compound interest rate
per interest period
Therefore, the compound amount
n = the number of
would be $1,295.65.
compounding periods
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 62 of 78
Compound Interest - General
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 63 of 78
Compound Interest - General
EXAMPLE
(Quarterly Compound) Assume that a $500 investment earns interest compounded quarterly.
Express the value of the investment after one year as a polynomial in the annual rate of interest r.
SOLUTION
r

A  P 1  
 m
mt
 r
A  5001  
 4
 r
A  5001  
 4
© 2010 Pearson Education Inc.
 4 1
Since interest is not being
compounded annually, we must use
this formula.
Replace P with 500, m with 4 (interest
is compounded 4 times each year),
and t with 1 (interest is being
compounded for 1 year).
4
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 64 of 78
§ 0.6
Functions and Graphs in Application
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 65 of 78
Section Outline

Geometric Problems

Cost, Revenue, and Profit

Surface Area

Functions and Graphs
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 66 of 78
Geometric Problems
EXAMPLE
(Fencing a Rectangular Corral) Consider a rectangular corral with two partitions, as shown below.
Assign letters to the outside dimensions of the corral. Write an equation expressing the fact that the
corral has a total area of 2500 square feet. Write an expression for the amount of fencing needed to
construct the corral (including both partitions).
SOLUTION
First we will assign letters to represent the dimensions of the corral.
y
x
x
x
x
y
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 67 of 78
Geometric Problems
CONTINUED
Now we write an equation expressing the fact that the corral has a total area of 2500 square feet.
Since the corral is a rectangle with outside dimensions x and y, the area of the corral is represented
by:
A  xy
Now we write an expression for the amount of fencing needed to construct the corral (including
both partitions). To determine how much fencing will be needed, we add together the lengths of all
the sides of the corral (including the partitions). This is represented by:
F  xxxx y y
F  4x  2 y
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 68 of 78
Cost Problems
EXAMPLE
(Cost of Fencing) Consider the corral of the last example. Suppose the fencing for the boundary of
the corral costs $10 per foot and the fencing for the inner partitions costs $8 per foot. Write an
expression for the total cost of the fencing.
SOLUTION
This is the diagram we drew to represent the corral.
y
x
x
x
x
y
Since the boundary of the fence is represented by the red part of the diagram, the length of fencing
for this portion of the corral is x + x + y + y = 2x + 2y. Therefore the cost of fencing the boundary of
the fence is (2x + 2y)(cost of boundary fencing per foot) = (2x + 2y)(10) = 20x + 20y.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 69 of 78
Cost Problems
CONTINUED
Since the inner partitions of the fence are represented by the blue part of the diagram, the length of
fencing for this portion of the corral is x + x = 2x. Therefore the cost of fencing the inner partitions
of the fence is (2x)(cost of inner partition fencing per foot) = (2x)(8) = 16x.
Therefore, an expression for the total cost of the fencing is:
(cost of boundary fencing) + (cost of inner partition fencing)
(20x + 20y) + (16x)
36x + 20y
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 70 of 78
Surface Area
EXAMPLE
Assign letters to the dimensions of the geometric box and then determine an expression representing
the surface area of the box.
SOLUTION
First we assign letters to represent the dimensions of the box.
z
y
x
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 71 of 78
Surface Area
CONTINUED
z
y
x
Now we determine an expression for the surface area of the box. Note, the box has 5 sides which
we will call Left (L), Right (R), Front (F), Back (B), and Bottom (Bo). We will find the area of each
side, one at a time, and then add them all up.
L: yz
R: yz
F: xz
B: xz
Bo: xy
Therefore, an expression that represents the surface area of the box is: yz + yz + xz + xz + xy =
2yz +2xz +xy.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 72 of 78
Cost, Revenue, & Profit
EXAMPLE
(Cost, Revenue, and Profit) An average sale at a small florist shop is $21, so the shop’s weekly
revenue function is R(x) = 21x where x is the number of sales in 1 week. The corresponding weekly
cost is C(x) = 9x + 800 dollars.
(a) What is the florist shop’s weekly profit function?
(b) How much profit is made when sales are at 120 per week?
(c) If the profit is $1000 for a week, what is the revenue for the week?
SOLUTION
(a) Since Profit = Revenue – Cost, the profit function, P(x), would be:
P(x) = R(x) – C(x)
P(x) = 21x – (9x + 800)
P(x) = 21x – 9x - 800
P(x) = 12x - 800
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 73 of 78
Cost, Revenue, & Profit
CONTINUED
(b) Since x represents the number of sales in one week, to determine how much profit is made when
sales are at 120 per week, we will replace x with 120 in the profit function and then evaluate.
P(120) = 12(120) - 800
P(120) = 1,440 - 800
P(120) = 640
Therefore, when sales are at 120 per week, profit is $640 for that week.
(c) To determine the revenue for the week when the profit is $1000 for that week, we use an
equation that contains profit, namely the profit function:
P(x) = 12x - 800
Now we replace P(x) with 1000 and solve for x.
1000 = 12x - 800
1800 = 12x
150 = x
Therefore x, the number of units sold in a week, is 150 when profit is $1000.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 74 of 78
Cost, Revenue, & Profit
CONTINUED
Now, to determine the corresponding revenue, we replace x with 150 in the revenue function.
R(x) = 21x
R(150) = 21(150)
R(150) = 3,150
Therefore, when profit is $1000 in a week, the corresponding revenue is $3,150.
NOTE: In order to determine the desired revenue value in part (c), we needed to solve for R(x).
But in order to do that, we needed to have a value for x to plug into the R(x) function. In order to
acquire that value for x, we needed to use the given information – profit is $1000.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 75 of 78
Functions & Graphs
EXAMPLE
The function f (r) gives the cost (in cents) of constructing a 100-cubic-inch cylinder of radius r
inches. The graph of f (r) is given.
(a) What is the cost of constructing a cylinder of radius 6 inches?
(b) Interpret the fact that the point (3, 162) is on the graph of the function.
(c) Interpret the fact that the point (3, 162) is the lowest point on the graph of the function. What
does this say in terms of cost versus radius?
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 76 of 78
Functions & Graphs
CONTINUED
SOLUTION
To determine the cost of constructing a cylinder of radius 6 inches, we look on the graph where r
= 6. The corresponding y value will be the cost we are seeking.
The red arrow is emphasizing the point in which we are interested. The y value of that point is
270. Therefore, the cost of constructing a cylinder of radius 6 inches is 270 cents or $2.70.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 77 of 78
Functions & Graphs
CONTINUED
(b) The fact that the point (3, 162) is on the graph tells us that the cost to make 100-cubic-inch
cylinders with a radius as small as 3 inches is 162 cents or $1.62.
(c) The fact that the point (3, 162) is the lowest point on the graph tells us that the least expensive
100-cubic-inch cylinder that can be made is a 3 inch cylinder at a cost of $1.62. Therefore, the 3
inch cylinder is the most cost-effective one that is offered.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 78 of 78