Download Chapter 2 Functions and Graphs - Business Mathematics & Statistics

INTRODUCTORY MATHEMATICAL ANALYSIS
For Business, Economics, and the Life and Social Sciences
Chapter 2
Functions and Graphs
2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics
2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL ANALYSIS
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
Chapter Objectives
• To understand what functions and domains are.
• To introduce different types of functions.
• To introduce addition, subtraction, multiplication,
division, and multiplication by a constant.
• To introduce inverse functions and properties.
• To graph equations and functions.
• To study symmetry about the x- and y-axis.
• To be familiar with shapes of the graphs of six
basic functions.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
Chapter Outline
2.1) Functions
2.2) Special Functions
2.3) Combinations of Functions
2.4) Inverse Functions
2.5) Graphs in Rectangular Coordinates
2.6) Symmetry
2.7) Translations and Reflections
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.1 Functions
• A function assigns each input number to one
output number.
• The set of all input numbers is the domain of
the function.
• The set of all output numbers is the range.
Equality of Functions
• Two functions f and g are equal (f = g):
1. Domain of f = domain of g;
2. f(x) = g(x).
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.1 Functions
Example 1 – Determining Equality of Functions
Determine which of the following functions are equal.
( x  2)( x  1)
a. f ( x ) 
( x  1)
b. g ( x )  x  2
 x  2 if x  1
c. h( x )  
 0 if x  1
 x  2 if x  1
d. k ( x )  
 3 if x  1
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.1 Functions
Example 1 – Determining Equality of Functions
Solution:
When x = 1,
f 1  g 1 ,
f 1  h1 ,
f 1  k 1
By definition, g(x) = h(x) = k(x) for all x  1.
Since g(1) = 3, h(1) = 0 and k(1) = 3, we conclude
that
g  k,
g  h,
hk
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.1 Functions
Example 3 – Finding Domain and Function Values
Let g ( x)  3x 2  x  5 . Any real number can be used
for x, so the domain of g is all real numbers.
a. Find g(z).
Solution: g ( z )  3z 2  z  5
b. Find g(r2).
Solution: g (r 2 )  3(r 2 )2  r 2  5  3r 4  r 2  5
c. Find g(x + h).
Solution: g ( x  h)  3( x  h) 2  ( x  h)  5
 3 x 2  6hx  3h 2  x  h  5
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.1 Functions
Example 5 – Demand Function
Suppose that the equation p = 100/q describes the
relationship between the price per unit p of a certain
product and the number of units q of the product that
consumers will buy (that is, demand) per week at the
stated price. Write the demand function.
100
Solution: q 
p
q
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.2 Special Functions
• We begin with constant function.
Example 1 – Constant Function
Let h(x) = 2. The domain of h is all real numbers.
h(10)  2
h(387)  2
h( x  3)  2
A function of the form h(x) = c, where c = constant, is
a constant function.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 3 – Rational Functions
x2  6 x
f ( x) 
x5
a.
is a rational function, since the
numerator and denominator are both polynomials.
b. g ( x)  2 x  3 is a rational function, since
2x  3 
2x  3
1
Example 5 – Absolute-Value Function
Absolute-value function is defined as x , e.g.
 x if x  0 
x 


x
if
x

0


2007 Pearson Education Asia
.
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 7 – Genetics
Two black pigs are bred and produce exactly five
offspring. It can be shown that the probability P that
exactly r of the offspring will be brown and the others
black is a function of r ,
r
5 r
1 3
5!   
4
4
P(r )     
r ! 5  r !
r  0,1, 2,...,5
On the right side, P represents the function rule. On
the left side, P represents the dependent variable.
The domain of P is all integers from 0 to 5, inclusive.
Find the probability that exactly three guinea pigs will
be brown.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.2 Special Functions
Example 7 – Genetic
Solution:
3
2
1 3
 1  9 
5!    120   
4
4
 64  16   45
P(3)      
3!2!
6(2)
512
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.3 Combinations of Functions
• We define the operations of function as:
( f  g )( x) 
(f  g )( x) 
( fg )( x) 
f
( x) 
g
f ( x)  g ( x)
f ( x)  g ( x)
f ( x).g ( x)
f ( x)
for g ( x)  0
g ( x)
Example 1 – Combining Functions
If f(x) = 3x − 1 and g(x) = x2 + 3x, find
2007 Pearson Education Asia
a. ( f  g )( x)
b. ( f  g )( x)
c. ( fg )( x)
f
d.
( x)
g
1
e. ( f )( x)
2
Chapter 2: Functions and Graphs
2.3 Combinations of Functions
Example 1 – Combining Functions
Solution:
a. ( f  g )( x)  f ( x)  g ( x)  (3 x  1)  ( x 2 +3x)  x 2  6 x  1
b. ( f  g )( x)  f ( x)  g ( x)  (3 x  1)  ( x 2 +3x)  1  x 2
c. ( fg )( x)  f ( x) g ( x)  (3 x  1)( x 2  3 x)  3 x 3  8 x 2  3 x
f
f ( x) 3x  1
d.
( x) 
 2
g
g ( x) x  3x
1
1
1
3x  1
e. ( f )( x)  ( f ( x))  (3 x  1) 
2
2
2
2
Composition
• Composite of f with g is defined by ( f g )( x)  f ( g ( x))
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.3 Combinations of Functions
Example 3 – Composition
If F ( p)  p 2  4 p  3, G ( p )  2 p  1, and H ( p )  p ,find
a. F (G ( p))
b. F (G ( H ( p )))
c. G ( F (1))
Solution:
a. F (G ( p))  F (2 p  1)  (2 p  1) 2  4(2 p  1)  3  4 p 2  12 p  2  ( F G )( p)
b. F (G ( H ( p)))  ( F (G H ))( p)  (( F G ) H )( p)  ( F G )( H ( p)) 
2
( F G )( p )  4 p  12 p  2  4 p 2  12 p  2
c. G ( F (1))  G (12  4  1  3)  G (2)  2  2  1  5
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.4 Inverse Functions
• An inverse function is defined as f ( f 1 ( x))  x  f 1 ( f ( x))
Example 1 – Inverses of Linear Functions
Show that a linear function is one-to-one. Find the
inverse of f(x) = ax + b and show that it is also linear.
Solution:
Assume that f(u) = f(v), thus au  b  av  b .
We can prove the relationship,
x b
( f g )( x)  f ( g ( x))  a
 b  ( x  b)  b  x
a
(g
(ax  b)  b ax
f )( x)  g ( f ( x)) 

x
a
a
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.4 Inverse Functions
Example 3 – Inverses Used to Solve Equations
Many equations take the form f(x) = 0, where f is a
function. If f is a one-to-one function, then the
equation has x = f −1(0) as its unique solution.
Solution:
Applying f −1 to both sides gives f 1  f  x    f 1  0.
1
1
f
(
f
(0))

0
f
(0) is a solution.
Since
,
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.4 Inverse Functions
Example 5 – Finding the Inverse of a Function
To find the inverse of a one-to-one function f , solve
the equation y = f(x) for x in terms of y obtaining x =
g(y). Then f−1(x)=g(x). To illustrate, find f−1(x) if
f(x)=(x − 1)2, for x ≥ 1.
Solution:
Let y = (x − 1)2, for x ≥ 1. Then x − 1 = √y and hence
x = √y + 1. It follows that f−1(x) = √x + 1.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
• The rectangular coordinate system provides a
geometric way to graph equations in two
variables.
• An x-intercept is a point where the graph
intersects the x-axis. Y-intercept is vice versa.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 1 – Intercepts and Graph
Find the x- and y-intercepts of the graph of y = 2x + 3,
and sketch the graph.
Solution:
3
When y = 0, we have 0  2 x  3 so that x  
2
When x = 0, y  2(0)  3  3
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 3 – Intercepts and Graph
Determine the intercepts of the graph of x = 3, and
sketch the graph.
Solution:
There is no y-intercept, because x cannot be 0.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.5 Graphs in Rectangular Coordinates
Example 7 – Graph of a Case-Defined Function
Graph the case-defined function
if 0  x < 3
 x

f ( x)   x  1 if 3  x  5
 4
if 5 < x  7

Solution:
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.6 Symmetry
• A graph is symmetric about the y-axis when (-a,
b) lies on the graph when (a, b) does.
Example 1 – y-Axis Symmetry
Use the preceding definition to show that the graph
of y = x2 is symmetric about the y-axis.
Solution:
When (a, b) is any point on the graph, b  a 2.
When (-a, b) is any point on the graph, (a)2  a 2  b.
The graph is symmetric about the y-axis.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.6 Symmetry
• Graph is symmetric about the x-axis when (x, -y)
lies on the graph when (x, y) does.
• Graph is symmetric about the origin when (−x,−y)
lies on the graph when (x, y) does.
• Summary:
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.6 Symmetry
Example 3 – Graphing with Intercepts and Symmetry
Test y = f (x) = 1− x4 for symmetry about the x-axis,
the y-axis, and the origin. Then find the intercepts
and sketch the graph.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.6 Symmetry
Example 3 – Graphing with Intercepts and Symmetry
Solution:
Replace y with –y, not equivalent to equation.
Replace x with –x, equivalent to equation.
Replace x with –x and y with –y, not equivalent to
equation.
Thus, it is only symmetric about the y-axis.
Intercept at 1  x 4  0
x  1 or x  1
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.6 Symmetry
• A graph is symmetric about the y = x when (b, a)
and (a, b).
Example 5 – Symmetry about the Line y = x
Show that x2 + y2 = 1 is symmetric about the line
y = x.
Solution:
Interchanging the roles of x and y produces
y2 + x2 = 1 (equivalent to x2 + y2 = 1).
It is symmetric about y = x.
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.7 Translations and Reflections
• 6 frequently used functions:
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.7 Translations and Reflections
• Basic types of transformation:
2007 Pearson Education Asia
Chapter 2: Functions and Graphs
2.7 Translations and Reflections
Example 1 – Horizontal Translation
Sketch the graph of y = (x − 1)3.
Solution:
2007 Pearson Education Asia