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Comments to the presenter.
 1. All presented problems are solved.
 2. Examples are suggested to be discussed,
problems to be solved by students.
 3. Check for the note “click”, please. If you see it, then there
are steps in discussion of the concept or the solution of the
problem.
 4. Any comments and suggestions to improve the session
would be appreciated.
Review Session
Basics of Functions & Their
Graphs
Pretest.
Click to check answers.
1. Domain?
Range?
2. Find
3. Find x for which
f(1) =
f(x) = -5
f(2) =
f(x) = 0
4. Identify the intervals over which the function is
increasing/decreasing/constant
Pretest - Solutions.
1. Domain?
Range?
2. Find
(, )
[5, )
3. Find x for which
f(1) = -4
f(x) = -5
f(2) = -5
for x = 2
f(x) = 0
4. Identify intervals over which the function is
increasing/decreasing/constant
f(x) is decreasing on (, 2) ,
increasing on (2, )
for x = -0.3 & 4.3
Learning Objectives for Part I
Functions, Domain & Range, Function Notation.








Apply the definition of a function.
Identify domain and range of a function.
Identify the domain & range from a graph.
Identify the domain & range from an equation.
Use function notation & evaluate a function.
Find the Difference Quotient.
Read graphs of functions.
Piecewise-Defined Functions.
Apply the definition of a function.
Definition of a Function
 What is a function? A correspondence (a rule that matches the
elements) between two sets D and R such that to each element of
the first set, D, there corresponds exactly one and only one
element of the second set, R.
 The first set is called the domain, and the set of corresponding
elements in the second set is called the range.
Domain: the set of all first entries (inputs) to be used in the rule (also
called domain elements or independent variables).
Range: the set of all second entries (outputs) that result from applying
the rule (also called range elements or dependent variables).
Independent Variable,
input
( x, y )
Dependent Variable,
output
Functions are commonly represented in 4
ways.
1.
2.
3.
Paycheck (P) depends on hours worked (h).
Verbally by a sentence.
Numerically by a list of ordered pairs or by a table.
F  (1,2),( 2,4),(3, 1)
Function
x
y
H  ( 4,1),(2,1),(2,0)
Not a function
–4
16
–2
4
0
0
2
4
4
16
Graphically by points on a graph
in a coordinate plane.
Function
Function
Use the definition of a
function to determine whether
each example is a function.
Definition of function:
To each element of the first
set, D, there corresponds one
and only one element of the
second set, R.
Click for each answer.
Not a function
4. Algebraically by an equation in two variables.
y  x2  4
Function
Function
y2  x
Not a function
Graphical Representation.
Use Vertical Line Test for Defining Functions.
If any possible vertical line intersects a graph in
at most one point, then the graph is a graph of a function.
Function.
Not a Function.
Identify domain and range of a function
from the graph.
Comment: In this presentation,
please assume that the graph
continues outside of the portion of
the coordinate system that is shown,
unless otherwise indicated.
Example
Graphical Representation.
Find the Domain & Range of the Function.
Domain: set of first components in the relation (inputs, independent variables)
Range: set of second components in the relation (outputs, dependent variables)
How to identify the domain :
(on click)
 Domain: Look at the x-axis. Considering all the points on
the graph, identify all the x-values included on the graph.
(Think about whether the graph extends further to the
left and right. Include all x-values that occur on the graph.)
How to identify the range:
(on click.)
 Range: Look at the y-axis. Considering all the points on
the graph, identify all the y-values included on the graph.
(Think about whether the graph extends further up and
down. Include all y-values that occur on the graph.)
Domain:
(, )
Range:
[5, )
Practice 1
Graphical Representation.
Find the Domain & Range of the Function.
Answers on clicks.
Function?
Yes
(Vertical Line Test)
Type of Function?
Absolute Value function:
Domain?
(, )
Range?
 ,3
Practice 2
Graphical Representation.
Find the Domain & Range of the Function.
Answers on clicks.
Function?
Yes
(Vertical Line Test)
Name?
Square Root function
Domain?
[0, )
Range?
[2, )
Practice 3
Graphical Representation.
Find the Domain & Range of the Function.
Answers on clicks.
Function?
Yes
(Vertical Line Test)
Type of Function?
Reciprocal function
y
Domain?
1
x
  ,0    0 ,  
Range?
  ,0    0 ,  
How to Identify the domain &
range from the equation
(Some Examples appear below)
Representation by Equation.
Finding the Domain & Range.
If a function is specified by an equation and the domain is not
indicated, we agree to assume that the domain is the set of all
real inputs that produce real outputs.
This set of all outputs is the range.
To Identify the Correct Domain:
Begin with All the Reals   ,   and exclude any values of x
that would result in either of these types of problems:
Type 1: A denominator that equals zero.
Type 2: A negative radicand under a radical with an even index (root).
Excluding the x-values that would result in these problems is
referred to as “restricting the domain”.
Example
Representation by Equation.
Finding the Domain & Range.
y  x4
Click for each answer.
Function?
Yes.
(By the definition of a function)
Name?
Linear function
(could be written: ax +by = c, where b is not zero)
Domain? {x | x is a real number}
or
(  ,  )
(Since x may be any real number.)
or (  ,  )
[The y-value can also be any real number, since y is always 4 more than x.]
Range? { y | y is a real number}
Note: For any Linear function, if no restriction is specifically stated on the
domain or range, both the Domain and Range will always be (  ,  ) .
Example
Representation by Equation.
Finding the Domain & Range.
1
y
3x  5
Click for each answer.
 This expression is defined for all real values of x except where the
denominator of the fraction is zero. This means that we begin with
all real numbers and then restrict the domain by excluding x = 5/3.
Domain:
x x  
5
3
or
  , 53    53 ,  
 The output values for the expression will be any real number
except 0. (Why?)
Because any fraction with a constant as the numerator cannot equal zero.
Range:
 y y  0 or  ,0    0 ,  
Example
Representation by Equation.
Find the Domain & Range.
y  3x  2
Domain?
Click for each answer.
Consider: Is there any value of x we cannot use? For example, what is y if x = 0?
y
3(0)  2 
2 Not a real number! We must Restrict the domain.
Ask: For what values of x is y defined? (Note that a square root is an even root. )
Answer: y is defined only when the radicand (3x-2) is equal to or greater than zero:
Using algebra, this is written:
3x –2 ≥ 0
This is the domain:
Range?
2
3
x  23
and solving for x, we find,
Domain: [ 23 , )
If x = , then y=0, and when x has a value greater than 23 , the radicand is a
positive number, so the square root would also be a positive number.
Range: [0, )
The range is y ≥ 0.
Representation by Equation.
Find the Domain & Range.
 Conclusion.
In general, the domain of a function excludes values that would cause
division by zero (Type 1) when the variable appears in a denominator
or that would result in
an even root of a negative number (Type 2) when the function contains
an even root [even index] of an expression that contains the variable.
Click to continue.
If Type 1,
y
1
3x  5
If Type 2,
y  3x  2
To find the domain,
1. Set any denominator equal to zero;
2. Solve the resulting equation, and
3. Exclude all these solutions from the set of all real numbers.
1. For any radical with even index, set the radicand greater than or equal to
2. Solve the resulting inequality, and
3. Use the solution set as the domain.
() zero;
Practice 4
Representation by Equation.
Find the Domain & Range.
y   2 x  3
Click for answers.
Set the Radicand  0
and solve:
2x  3  0
Domain: (, 32 ]
2x  3
3
(,0]
Range:
x
2
What is the process to follow to identify the Domain and Range?
Name? Square Root function (Note: even root)
To find the Domain:
This problem has no denominator, but it does contain a radical with an even root.
Set the radicand  0 and solve for x. The solution set is the domain of the function.
To find the Range:
1) Determine any values that y will not equal and exclude those from the range.
2) A graph may be helpful.
Practice 4 (Solutions)
Representation by Equation.
Find the Domain & Range.
y   2 x  3
Name?
Click once for each answer.
Square Root function
Apply the Type 2 process since the radical has an even index (root).
Domain:
Set the radicand -2x+3 greater than or equal to zero  , and solve for x.
–2x + 3 > 0
Divide by –2 on each side and
–2x > –3
reverse (flip) the inequality symbol!
– 2x/(– 2) < – 3/(– 2)
x < 3/2
3
The domain is “ all real x < 3/2” or (, 2 ]
Range: For the range you may want to refer to the graph.
The range is “all real y < 0” or (,0]
Practice 5
Representation by Equation.
Find the Domain & Range.
x2  x  2
y 2
x x2
Name?
Domain:
Range:
Click for answers.
Rational function.
 x x  1, 2
(  ,  )
What is the process to follow to identify the Domain and Range?
To find the Domain: This function contains no radical, but there is a
denominator. Set the denominator to zero. Solve for x, and exclude any
resulting values of x from (  ,  ).
To find the Range for this function, a graph will be helpful.
Practice 5 (Solutions)
Representation by Equation.
Find the Domain & Range.
x2  x  2
y 2
x x2
Name?
Click once for each answer.
Rational function.
Domain:
Apply the Type 1 process to identify restrictions on the domain.
Set the denominator equal to zero, and solve. Exclude all
values of x that would result in a denominator value of 0:
x2 – x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 or x = –1
The domain is "all real x ≠ –1, 2” or  x x  1, 2
In interval form the domain is: (,  1)  (1, 2)  (2, )
Range: The range requires the graph.
As we see from the picture, the graph will eventually include or “cover” all
possible values of y, so the range is "all real numbers” or (  ,  ) .
Practice 6
Representation by Equation.
Find the Domain & Range.
( A) y   x4  4, ( B) y  - x3  2 x  1
Click for answers.
Name? Equations like the above represent Polynomial functions.
Domain:
(  ,  )
Range: ( A) (  ,4],
(B) (  ,  )
Range of (A): In the graph of (A) when x is zero, y is
4, and otherwise y < 4.
Range of (B): All real numbers will occur on the graph
of (B) as y-values.
Polynomial functions have no denominators (so no division-by-zero problems) and
no radicals (so no even-root-of-a-negative problems).
For any polynomial function, the domain is "all real values of x” (  ,  ) .
For any polynomial of odd degree, the range is always (  ,  ) .
The Range of a polynomial of even degree depends on the
sign of the leading coefficient: Positive Sign: [a,  ) ; Negative sign: ( , a ]
Where “a” represents some real number value.
Practice 6 (Solutions)
Representation by Equation.
Find the Domain & Range.
( A) y   x4  4, ( B) y  - x3  2 x  1
Click once for each answer.
Name? Equations like those shown above, where y equals a sum of terms in
which x has integer exponents represent
Polynomial functions.
Domain: (  ,  )
For any polynomial function, the domain is "all real values of x” ( ,  ) .
You may choose to sketch the graph of the
function and find the range from the graph.
Consider the graphs that are shown at the right.
Each function has a negative leading coefficient.
Degree of Equation (A): even
Degree of Equation (B): odd
Range: ( A) (  ,4], (B) (  ,  )
( A)
(B)
y
y








x








x













y   x4  4



y  - x3  2 x  1
Function Notation and Evaluation.
Function Notation
Start on click.
 The notation f (x) will now be used instead of the variable y.
 This is read as “ f of x” and f(x) is equivalent to the
y-coordinate of the function corresponding to a given x value.
The equation
2
y  x 2
can now be expressed as
f ( x)  x 2  2
We say, “y is a function of x” to emphasize that y depends on x.
We use the notation y = f (x) , which is function notation.
Name of the function
Defining expression (formula)
y  f ( x)  x 2  2
Value of the function (y or f(x))
Independent variable
Example
Find f (2) for each different function shown.
Check each problem on click.
a. Table.
x
f(x)
–4 16
–2 4
0
2
4
0
4
16
f (2)  4
Given a value of x,
b. Set of ordered pairs. (the first coordinate or
independent variable),
f = {(2, 6), (4, 2)}
f (2)  6
evaluate
the function f(x)
c. Graph.
using that x-value.
d. Equation.
f (2)  3
f(x) = –x2
f(2) = –(2)2
f (2)  4
Example
Function Evaluation
Start on click.
 Consider the function
f ( x)  x 2  2
It tells you the “instructions” you are to follow.
 What does f (-3) mean?
 Replace x with the value -3 and evaluate the expression:
f (3)  (3) 2  2
f (3)  11
 The notation means: “when x = -3, the value of f(x) is 11”
So the point (-3,11) is on the graph of function f.
Example
Evaluating a function
Given f(x) = x2 + 2x – 1, find f(2), f(–3), f(k).
Click for the answers.
Each location where you see an "x” in the function f(x) is just an open box, waiting for
something to be put into it.
f(2) = 7 since f(2) = (2)2 +2(2) – 1 = 7
f(–3) = 2 since f(–3) = (–3)2 +2(–3) – 1 = 2
f(k) = k2 + 2k – 1
Functions can be defined using a letter different from x. The function below is exactly
the same function as f(x) = x2 + 2x – 1. However, below it is defined using “m”.
f(m) = (m)2 + 2(m) – 1 = m 2 + 2 m – 1
You would still find f(2) or f(-3) using the same steps: in f(m) replace “m” with “2”
the first time and the second time replace the “m” with “-3”, just as was done above.
This is how to work with function notation .
Practice 7
Evaluate a Function
 Evaluate f(x) = 3x – 7 at x = 2, a, 3-x
Click once for the answers.
f(2) = -1
Click once for solutions.
f(2) = 3(2) – 7 = -1;
f(a) = 3a – 7
f(a) = 3(a) -7 = 3a – 7;
f(3 – x) = 2 – 3x
f(3 – x) = 3(3 – x) – 7 = 2 – 3x.
Difference Quotient
Example
Find the Difference Quotient
Click for steps.
Given f(x) = 3x2 + 2x, find
f ( x  h)  f ( x )
h
Note: Simplify the numerator
first. Put the “h” back in the
denominator later.
When working through more involved exercises like this example, start by splitting
the exercises into smaller steps, starting with the f(x + h) in the numerator:
 in f( ) = 3( )2 + 2( ) replace the x with (x+h) to find f(x + h):
f(x + h) = 3(x + h)2 + 2(x + h) = 3x2 + 6x h + 3 h 2 + 2x + 2h
 f(x) = 3x2 + 2x [The original function is repeated here.]


f(x + h) – f(x) = [3x2 + 6xh + 3h2 + 2x + 2h] – [3x2 + 2x]
= 3x2 + 6xh + 3h2 + 2x + 2h – 3x2 – 2x = 6xh + 3h2 + 2h
f ( x  h)  f ( x )
h

6 xh  3h 2  2h
h
 h(6 x  3h  2)  6x  3h  2
h
Practice 8
Find the Difference Quotient
Given f(x) = 2x2 - 2x + 1, find f ( x  h)  f ( x)
h
Click for the answer.
4x  2h  2
Click for the solution.
Practice 8 (Solution)
Find the Difference Quotient
Given f(x) = 2x2 - 2x + 1, find
f ( x  h)  f ( x )
h
1). f ( )  2( ) 2  2( )  1
Click once for solution.
2). f ( x  h)  2( x  h) 2  2( x  h)  1
 2( x 2  2 xh  h 2 )  2 x  2h  1
 2 x 2  4 xh  2h 2  2 x  2h  1
f ( x  h)  f ( x) 2 x 2  4 xh  2h 2  2 x  2h  1  (2 x 2  2 x  1)
3).

h
h
2 x 2  4 xh  2h 2  2 x  2h  1  2 x 2  2 x  1
4).

h
4 xh  2h 2  2h h(4 x  2h  2)
5).


h
h
 4 x  2h  2
Reading Graphs
Increasing/Decreasing/Constant
Functions
For points (x1, y1) and (x2, y2) on the graph,
 Increasing: The graph goes “up” as you move from left to right.
x1  x2 implies y1  y2
 Decreasing: The graph goes “down” as you move from left to
right.
x1  x2 implies y1  y2
 Constant: Graph remains horizontal as you move from left to right.
x1  x2 implies y1  y2
Example
Use the graph to determine the following.
Check each problem on click.
Yes
1. Is it a function?
2. Domain:
[0, 4]
Range:
(0, 2]
3. Find
5. Intervals on which f is
increasing, decreasing,
or constant.
f is increasing (up) on (2, 3),
decreasing (down) on  0,1   3,4 ,
constant (horizontal) on (1, 2).
(Vertical Line Test)
4. Find x for which
f(1) = 1
f(x) = 2
f(2) = 2
for x = 2, 3
f(4) = 0.5
f(x) = 1
f(5) =
undefined
for x =0 or x from [1, 2)
Piecewise-Defined Function.
Example
Evaluate Piecewise-Defined Function
2  2 x if x  2
f ( x)  
 x  2 if x  2
(I)
(II)
Evaluate f(0), f(2), f(3)
for the function
A piecewise-defined function is a function whose definition
involves more than one formula (has two or more “pieces”).
Notice that this function is defined by different rules I and II for different parts of its domain.
Click for each evaluation step.
f(0)=? Since 0 < 2,
use
(I)
f(0) = 2 –2(0) = 2,
f(2)=? Since 2 ≥ 2,
use
(II)
f(2) = (2) – 2 = 0,
f(3)= ? Since 3 ≥ 2,
use
(II)
f(3) = (3) – 2 = 1.
 Graphing such functions involves graphing each rule over the
appropriate portion of the domain on the same coordinate plane.
Example
Graph
Click to complete each
T-Chart (table) below.
(I)
y
x
-2
6
2
0
2
0
3
1
4
2
1
(I)
(II)
(II)
x
0
2  2 x if x  2
f ( x)  
 x  2 if x  2
y
(2) (-2)
Since the function has two pieces (rules) , we use two T-charts.
The break between the two “pieces" of this function
(the point at which the function changes rules) is at x = 2, so that
is where T-charts will break . The function is graphed in two
pieces: the first piece includes only x-values before x = 2, and
the second piece has domain from x = 2 to positive infinity.
Why is the last point in T-chart I in parentheses?
Click for each part of the graph.
Click for the answer.
Actually, x = 2 is not a domain value for T-chart I, but it is
helpful to know the y-value that the piece "approaches"
close to this endpoint of its domain (as x gets closer to 2).
Notice that the point (2,-2) is not included in the graph of
part I (shown with an open circle), but the point (2,0) is
included in graph of part II (shown with a closed circle).
(I)
(II)
Practice 9
Graph
 x 2  2 if x  1
f ( x)  
2 x  4 if x  1
(I)
(II)
Click for the graph.
(I)
(II)
Practice 9 (Solution)
Graph
(I)
 x 2  2 if x  1
f ( x)  
2 x  4 if x  1
(I)
(II)
x
y
(II)
-3
7
x
-2
2
1
1
Why would you need two T-charts for the graph? There are two pieces.
-1
-1
2
0
Why is the point (1,-1) in T-chart I in parentheses? It’s not on the graph.
0
-2
3
-2
Answer the questions.
y
(1) (-1)
Describe what happens to the graph of f(x) at x = 1?
The rule for the function changes. (This graph has a break or jump at x=1.)
Click for the graph.
Answer the questions.
The endpoint (1,-1) on the graph of piece I is an open
circle. (Why?). This point is not on the graph. 1 is not in
(I)
the domain for this piece of the graph.
The endpoint (1,2) of piece II is a closed circle. (Why?).
The point (1, 2) is on this piece of the graph. 1 is
included in the domain for this piece of the graph
(II)
Graph Transformations.
Learning Objectives for Part II
Graph Transformations.
The basic graph transformations are:
 Vertical and Horizontal translations (shifts),
 Vertical and Horizontal stretches and shrinks.
 Reflection in the x-axis and y-axis,
Vertical and Horizontal Translations of Function f(x)
g(x) = f(x) ± k (k > 0)
and
Start on click.
Start on click.
 Vertical shifts :
•
g(x) = f(x ± k) (k > 0)
y  x2  2
Move the graph up (+k) or down (-k);
•
Impact only the “y” values;
•
No changes are made to the “x” values.
 Horizontal shifts: y  ( x  2)2
•
Move the graph left (+k) or right (-k)
(HINT: go the opposite direction);
•
Impact only the “x” values;
•
No changes are made to the “y” values.
y  x2  2
yx
2
y  x2  2
y  ( x  2)2
yx
2
y  ( x  2)2
Practice 10 (with solution)
Combining a Vertical & Horizontal shifts
Graph: g ( x)  x  2  4
Click for the graph.
Click for each answer.
y
What is the basic function?
Absolute Value function

f ( x)  x
f ( x)  x



List the transformations to the basic function.
x

1. Horizontal shift right 2 units,
2. Vertical shift down 4 units.












g ( x)  x  2  4
Vertical and Horizontal Stretches and Compressions of
Function f(x)
g(x) = kf(x) (k >0)
or
g(x) = f(kx) (k >0)
Start on click.
Start on click.
 Vertical Stretches & Shrinks
 Horizontal Stretches & Shrinks
•
Stretch graph vertically (k > 1), or
Compress graph vertically (0< k < 1).
•
Stretch graph horizontally (0< k < 1),
Compress graph horizontally (k > 1).
•
Impact only the “y” values;
•
Impact only the “x” values;
•
No changes are made to the “x” values.
•
No changes are made to the “y” values.
Practice 11 (with solution)
y  3| x |
Graph: y  3 | x |
Basic function:
Click once for solution.
y  | x|
Transformation of the Graph:
Vertical Stretching (by factor of 3)
y | x |
y | x |
Graph:
Click once for solution.
y
1
|x|
4
y
Transformation of the Graph:
Vertical Compression (by factor of 1/4)
1
|x|
4
Reflections of Function f(x) around y-axis and x-axis
g(x) = -f(x)
or
g(x) = f(-x)
g(x) = -f(x)
 Reflection in the (around the) x-axis
g(x) = f(-x)
 Reflection in the (around the) y-axis
•
-f(x) reflects f(x) around the x-axis.
•
f(-x) reflects f(x) around the y-axis.
•
Impacts only the sign of the “y” values,
(e.g., positive y values become negative);
•
•
No changes are made to the “x” values.
Impacts only the sign of the “x” values,
(e.g., negative x values are changed to
positive);
•
No changes are made to the “y” values.
Click once to start.
Click once to start.
y x
y  x
y x
y x
Summary of Graph Transformations
 Reflection in y-axis: y = f (-x)
Reflect the graph of y = f (x) in the y-axis.
 Horizontal Stretch or Shrink: y = f (ax) (a >0)
 a > 1: Stretch graph of y = f (x) horizontally by multiplying each x-value by a.
 0< a <1: Shrink graph of y = f (x) horizontally by multiplying each x-value by a.
 Horizontal Shift: y = f (x ± c) (c >0)
 + c Shift graph of y = f (x) left c units.
 - c Shift graph of y = f (x) right c units.
 Vertical Stretch or Shrink: y = af (x) (a >0)
 a > 1: Stretch graph of y = f (x) vertically by multiplying each ordinate value by a.
 0< a <1: Shrink graph of y = f (x) vertically by multiplying each ordinate value by a.
 Reflection in x-axis: y = -f (x)
Reflect the graph of y = f (x) in the x-axis.
 Vertical Shift: y = f (x) ± d (d >0)
 + d Shift graph of y = f (x) up d units.
 - d Shift graph of y = f (x) down d units.
Multiple Transformations
This would be the order of the
steps to follow if all these types
of transformations are needed:






Reflect in the y-axis
Horizontal Stretch or Shrink
Horizontal Shift
Vertical Stretch or Shrink
Reflect in the x-axis
Vertical Shift
However, the examples
included in this presentation
require only these steps:
1.
2.
3.
4.
Horizontal Shift
Vertical Stretch or Shrink
Reflection in the x-axis
Vertical Shift
Example
Show sequence of transformations of f(x) to get h( x )  2 x  4  2
State the domain and range of h(x).
Click for each step.
Basic function:
f ( x)  x
List of transformations:
f ( x)  x
h( x)  2 x  4  2
Domain h(x):
[4, )
Range h(x): (, 2]
1.
Horizontal shift:
4 units left;
2.
Vertical stretch:
by factor 2;
3.
Reflection:
around x-axis;
4.
Vertical shift:
2 units down.
Example
Graph using key points: h( x )  2 x  4  2
Click for each step.
Basic function:
f ( x)  x
Key points:
(0,0), (1,1), (4,2)
List of transformations of key points:
f ( x)  x
1.
Horizontal shift 4 units left
(subtract 4 from each x): (-4,0), (-3, 1), (0,2);
2.
Vertical Stretch by factor 2
(2 times all the y’s):
3.
(-4,0), (-3,2), (0,4);
Reflection in the x-axis
(change signs of all y’s): (-4,0), (-3,-2), (0,-4);
4.
Vertical shift: 2 units down
(subtract 2 from each y): (-4,-2),(-3,-4),(0,-6).
h( x)  2 x  4  2
Sequence of transformations
Practice 10
Graph: g ( x ) = - 4( x - 3)3 - 1
Basic function: y= x3
Key points: (0,0), (1,1), (-1,-1)
List of transformations of key points:
f ( x)  4( x  3)3  1
f ( x)  x
1.
3
Horizontal shift 3 units right
(add 3 to each x):
2.
Vertical Stretch by factor 4:
(4 times all the y’s):
3.
(3,0), (4,4), (2,-4);
Reflection in the (around) x-axis:
(change signs of all y’s):
4.
(3,0), (4, 1), (2,-1);
(3,0), (4,-4), (2,4);
Vertical shift: 1 unit down:
(add 1 to each y):
(3,-1), (4, -5), (2,3).
Practice 10 (Solution)
Sequence of transformations
Graph: g( x ) = - 3( x - 2)2 + 1
Click for each step.
Basic function:
f ( x)  x 2
Key points:
f ( x)  x 2
(0,0), (1,1), (-1,1)
List of transformations of key points:
1.
Horizontal shift 2 units right
(add 2 to each x):
2.
Vertical Stretch by factor 3:
(3 times all the y’s):
3.
4.
(2,0), (3,3), (1,3);
Reflection in the (around) x-axis:
(change signs of all y’s):
g ( x)  3( x  2)2  1
(2,0), (3, 1), (1,1);
(2,0), (3,-3), (1,-3);
Vertical shift: 1 unit up:
(add 1 to each y):
(2,1), (3, -2), (1,-2).
Learning Objectives for Part III
Function Operations and Composition
 Arithmetic Operations on Functions and Domain.
 Composition of Functions and Domain.
 Inverse Functions.
Arithmetic Operations on Functions
and Domain.
Arithmetic Operations on Functions
Given two functions  and g, and for all x common to both
domains, the sum, difference, product & quotient of  and g are
defined as follows.
1. Sum:
( f  g )( x)  f ( x)  g ( x)
2. Difference:
( f  g )( x)  f ( x)  g ( x)
3. Product:
( fg )( x)  f ( x)  g ( x)
4. Quotient:
 f 
f ( x)
, g( x)  0
  ( x) 
g ( x)
g
The domain of an arithmetic combination of functions f and g consists of
all real numbers that are common to the domains of f and g.
For the quotient f(x)/g(x), there’s further restriction that g(x) ≠ 0.
Practice 11
Using Operations on Functions, Evaluating New Functions.
Let f ( x)  2 x  1 and g ( x)  x 2  2 x  1
f
Find (A) ( f  g )( x), ( f  g )(1); (B) ( f  g )( x), ( f  g )(2); (C) ( f  g )( x); (D)   ( x).
g
Solutions
Click for each answer.
( A) ( f  g )( x)  f ( x)  g ( x) = ( 2 x  1) +( x 2  2 x  1)  x 2  4 x,
( f  g )(1)  (1)2  4(1)  3.
( B) ( f  g )( x)  f ( x)  g ( x) = (2 x  1) - ( x 2  2 x  1)   x 2  2,
( f  g )(2)  (2)2  2  2;
(C ) ( fg )( x)  f ( x)  g ( x)  ( 2 x  1)( x 2  2 x  1)  2 x3  5 x 2  1
f 
f ( x)
2x 1
( D)   ( x) 
= 2
,
g ( x) x  2 x  1
g
x2  2x 1  0
The Domains
For functions  and g, the domains of  + g,  – g, and g
include all real numbers in the intersection (common part)
of the domains of  and g,
while the domain of /g includes those real numbers in the
intersection of the domains of  and g for which g(x) ≠ 0.
The two rules:
(1) Never allow a negative radicand under a radical sign with an
even index (root).
(2) No denominator of a fraction can be allowed to equal zero.
Example
Finding the Domains of Quotients of Functions
Click for steps.
f
g
Given f ( x) = x and g( x) = 4  x, find the domains of   ( x) and   ( x).
g
f
 f 
f ( x)
x
(
x
)

=
, g ( x)  0,
 
g ( x)
4 x
g
g
g ( x)
4 x
(
x
)

=
, f ( x)  0
 
f ( x)
x
 f 


Domain of  gf  ( x) : Since no square root can
have a negative radicand, and no denominator
can be zero, this domain can be found by
solving the inequalities x  0 and 4  x  0
and finding the intersection of the two
 f 
solution sets. The domain of  g  ( x) is [0,4).
Remember that any restrictions on the domains of f or g must be considered to
determine the domain when performing arithmetic operations on functions.
The domain of f is [0,  ) (why?), and the domain of g is (, 4] (why?)
The intersection (common part) of these domains is
[0, 4]
(refer to a number line).
For a quotient, we also have the restriction that the denominator cannot be 0.
So, the domain of (f/g) is [0, 4)
,
and the domain of (g/f) is (0, 4] .
Why are the two domains different? In f/g x=4 would give a 0 denominator as would x=0 in g/f .
Practice 12 (with answers)
Using Operations on Functions, Finding Domains.
Let f ( x)  8 x  9 and g ( x)  2 x  1
f
Find (A) ( f  g )( x), ( f  g )( x), ( f  g )( x),   ( x), and (B) their domains.
g

Click once for the solutions to (A).
( A) ( f  g )( x)  f ( x)  g ( x) = 8 x  9  2 x  1,
( f  g )( x)  f ( x)  g ( x) = 8 x  9  2 x  1,
( fg )( x)  f ( x) g ( x) = (8 x  9) 2 x  1,
Click again for the answers to (B).
f 
f ( x) 8 x  9
(
x
)

=
.
 
g ( x)
2x 1
g
(B) The domains of  + g,  – g, g:
The domain /g:
 21 , 
( 21 , )
See next slide for the solutions to (B) .
Practice 12 (solutions for B)
Finding Domains.
f ( x)  8 x  9 and g ( x)  2 x  1
Click for steps.
1) Find the domains of  and g.
The domain of  is the set of all real numbers: (– , );
The domain of g is the set of all real numbers that make 2x – 1 nonnegative:
 21 , 
2) The domain each of these functions is the intersection of the domains of  and g:
The domain for each of  + g,  – g, g is:
 ,     21 ,     21 ,  
 f 
8x  9
Domain of Numerator: (, )
3)   ( x) =
The intersection is  21 ,   .
1
2 x  1 Domain of Denominator: [ 2 , )
g
But no denominator can be equal to zero so, g ( x)  2 x  1  0 which means
1
2x 1  0 so we know x  2
The domain of /g : ( 21 , )
Composition of Functions
and its Domain.
Composition of Functions
and its Domain.
If  and g are functions, then the composite
function, or composition of g and , is defined by
 g f  x   g  f ( x)
The domain of g f is the set of all real numbers x
in the domain of  such that (x) is in the domain of g.
Example
Composition of Functions
Given f ( x)  x 2 and g ( x)  x  1,
( A) Evaluate ( g f )(3);
( B) Find 1) ( g f )( x), 2) ( f g )( x).
Click for steps.
(A)
( g f )(3)  g[ f (3)]  g (9)  10
 f g  3   10
(B) 1) ( g f )( x)  g[ f ( x)]  g ( x 2 )  x 2  1
( g f )( x)  x 2  1
2) ( f g )( x)  f [ g ( x)]  g ( x  1)  ( x  1)2  x 2  2 x  1
( f g )( x)  x 2  2 x  1
Example
(
( f  g )(0),
( gf )(1),
(
f
)(0),
g
g
)(0),
f
( f g )(1),
( g f )(1).
undefined
0
2
0.5
11.5
Click for answers (answer color matches problem color).
2.8 - 71
Example
Composition of Functions
and Domain.
6
1
Given f ( x ) 
and g ( x )  , find (A) ( g
x3
x
f )(x ) and (B) its domain.
Click for steps.
( A) ( g f )( x )  g  f(x)  g 
(B) Domain?
6
x 3

1
6
x 3
x3

6
x3
( g f )( x ) 
6
Begin with the domain of f , which is all real numbers except 3. Exclude any values of x
for which f(x) is not in the domain of function g and any other values of x for which
( g f )(x) would be undefined. Since 3 is the only x-value that needs to be excluded, the
domain of the function is all reals except 3.
Would it be correct to say that the domain for this composition
function is the set of all real numbers???
Example (cont.)
Composition of Functions
and Domain.
6
1
Given f ( x ) 
and g ( x )  , find (A) ( g
x3
x
f )(x ) and (B) its domain.
Click for steps.
1). The domain of  is all real numbers except 3. (why?)
2). The domain of g is all real numbers except 0. (why?) Will f(x)= 0 for any x-value?
( g f )( x)  g  f(x)
Input for g(x)
3).The function values of f(x) are used as “x” in g. Will the denominator of g equal 0?
4). For the domain of g f , begin with the domain of the first function applied,
function f. Exclude values of x where f(x) is not in the domain of function g and any other
values of x for which ( g f )(x) would be undefined. For this example, only the number 3 must
be excluded from the set of all real numbers.
The domain of g f is  x x  3 or  ,3    3 ,  
Practice 13
Composition of Functions
and Domain.
6
1
Given f ( x ) 
and g ( x )  , find (A) ( f
x3
x
g )(x ) and (B) its domain.
Click for answers.
6
6
6x
 1
( A) : ( f g )( x )  f [g ( x )]  f   


 x  1  3 1 3x 1 3x
x
x
 1  1 
(B ) :  ,0    0,    ,  
 3 3 
Begin with the domain of function g, which is all real values of x except zero.
Exclude any values of x for which g(x) is not in the domain of function f and any other
values of x for which ( f g )(x )would be undefined.
Exclude both 0 and 1/3 to arrive at the domain (B) of the composite function.
(Note: Continue for additional information concerning deriving this composite function.)
Practice 13 (Solution for A)
Composition of Functions.
6
1
Given f ( x ) 
and g ( x )  , find (A) ( f
x3
x
g )(x ) and (B) its domain.
Click for steps.
6
 1
( A) ( f g )( x )  f(g(x))  f   
 x 13
x
6x
( f g )( x ) 
1 3 x
To simplify the complex
fraction multiply both the
numerator and denominator
by x.
Or change the denominator
to fraction form and then
multiply by its reciprocal.
Practice 13 (Solution for B)
Composition of Functions - Domain.
Given f ( x ) 
6
1
and g ( x )  , find (A) ( f
x3
x
g )(x ) and (B) its domain.
Click for steps.
(B) Domain: 1). The domain of  is all real numbers except 3. (why?)
2). The domain of g is all real numbers except 0. (why?)
3). ( f g )( x)  f  g  x  
Input for f(x)
4). g(x) = 3 for some x. However, 3 cannot be used in function f;
Then solve g(x) = 3 and exclude the solution from the domain of f g.
1
x
3
1  3x
x  31
5).Therefore the domain of f g is the set of all
real numbers except 0 and 31:
 1  1 
 ,0    0,    ,  
3
3




Inverse Functions.
Inverse Functions
If for any value x, ( f
g )( x )  x and (g
f )( x )  x,
then g is called "the inverse of function f" (f 1 or "f - inverse" ).
For a function  to have an inverse,  must be a one-to-one function.
A function is said to be one-to-one if each range value corresponds
to exactly one domain value.
By the definition of inverse function, the domain of  is the range of -1,
and the range of  is the domain of -1
Note: Any function that is either increasing
or decreasing on its entire domain has an
inverse function.
Example
Decide if the Given Functions are Inverses.
Let f ( x )  x 3 -1 and g ( x )  3 x  1.
Steps
Is g the inverse function of f ?
1) Apply the Horizontal Line Test to graphs to see if the functions are one-to-one.
2) Check to see if both statements are true: a) ( ◦ g)(x) = x and b) (g ◦ )(x) = x.
Click for steps.
1) Both functions are one-to-one. (Why? - give your reason)
2) a) ( ◦ g)(x) = f(g(x))
 f

3

x 1
b) (g ◦ )(x) =
Prove on your own.
 g  x3  1
 ( x  1) -1
 3 ( x3  1)  1
 x  1 1  x

3
3
3
Check the proof.
x3  x
Conclusion: Functions f and g are inverse functions. (g = f -1)
Important Facts About Inverses
Click once)
1. If  is one-to-one, then -1 exists. (For graphs use the Horizontal Line Test)
2. The domain of  is the range of -1, and
the range of  is the domain of -1.
3. If the point (a, b) lies on the graph of , then (b, a) lies on the graph of
-1, so the graphs of  and -1 are reflections of each other across the
line y = x.
4. To find the equation for -1,
1) replace (x) with y;
2) interchange x and y;
3) solve for y;
4) replace y with -1. This is your candidate for the inverse function.
5) prove that ( º-1)(x)=x and (-1 º)(x)=x
6) Celebrate!
Example
FINDING THE INVERSE OF A FUNCTION WITH A
RESTRICTED DOMAIN
1
Let f ( x)  x  5, find f ( x).
First, notice that the domain of  is restricted to the interval [– 5, ).
Function  is one-to-one because it is increasing on its entire domain
and, thus, has an inverse function. Apply steps.
Click for each step.
f ( x)  x  5,
y  x  5,
x  y  5,
1) replace (x) with y;
2) interchange x and y;
3) solve for y;
2
x  y 5 ,
2
y  x2  5
f 1 ( x)  x 2  5
4) replace y with -1
Continued on next slide.
Example (cont.)
FINDING THE INVERSE OF A FUNCTION WITH A
RESTRICTED DOMAIN
However, we cannot define -1 as x2 – 5.
The domain of  is [–5, ), and range is [0, ).
The domain of -1 must match the range of  , so we define the domain
of -1 as shown in red here: f 1 ( x)  x 2  5, x  0
Click to continue.
5) Prove that ( º-1)(x)=x and (-1 º)(x)=x on your
own. [If the result is “x” both times, it proves that
the two functions are inverses functions.]
As a check, graphs of  and -1 are helpful.
The domain of -1, [0, ), is the range of .
The line y = x is included on the graphs to show that
the graphs are mirror images with respect to this line.
Practice 14
Finding the Inverse of a
Function
Given f ( x)  x 2  2
decide whether the equation defines a
one-to-one function. If so, find the equation of the inverse.
Click for the answer.
The given equation does not define a one-to-one function.
The given function does not have an inverse on its current domain.
However, if the domain were restricted so that the function would then
be a one-to-one function, the function with restricted domain would
then have an inverse. If the domain were restricted to the interval
[0, ) , then the function would have an inverse on this restricted
domain. The equation of the inverse function would be
f
1
( x) 
x2
Post-test.
Problem 1
Find the Domain & Range.
1
y
x4
2
Problem 2
Find the Domain & Range.
5
y
x 1
Problem 3
Graph the Function Using Sequence of
Transformations
g ( x)  3( x  2)3  1
Problem 4
Find the Inverse of a Function.
Given f ( x)  x 2 +2, x  0 find f 1 ( x).
Problem 1(Solutions)
Find the Domain & Range.
1
y
x4
2
Click for each answer.
Domain: [8,  )
Set the radicand to be greater than or equal to zero and solve the resulting inequality:
1
x40
2
x 8  0
Range: [0, )
Multiply both sides by 2:
Add 8 to both sides:
x 8
Since y=0 when x=8 and the radicand is non-negative, for all x-values greater
than or equal to 8, the range is y ≥ 0.
Problem 2 (Solutions)
Find the Domain & Range.
5
y
x 1
Click for each solution.
Domain:
 ,1  1,  
The domain is all the values that x can equal.
The domain for this function includes all real numbers except 1.
Range:
 ,0    0 ,  
The range: all real numbers except 0.
(No function with a constant numerator can be equal to zero.)
Problem 3 (Solution)
Graph the Function Using Sequence of
Transformations
Basic function:
g ( x)  3( x  2)  1
3
Key-points:
f ( x)  x 3
(-1,-1), (0,0), (1,1)
y

f ( x)  x

1st transformation would be Horizontal shift:
2 units left (subtract 2 from each x). The
points are now: (-3,-1), (-2,0), (-1,1)
3

g ( x)  3( x  2) 1
3

x









2nd transformation would be Vertical stretch:
3 times all the y’s. The points are now:
(-3,-3), (-2,0), (-1,3)



3rd transformation would be Vertical shift:
1 unit down (subtract 1 from all y’s). The
points are now: (-3,-4), (-2,-1), (-1,2)
Problem 4 (Solution)
Find the Inverse of a Function.
Given f ( x)  x 2 +2, x  0 find f 1 ( x).
First, notice that the domain of  is restricted to the interval [0, ).
Function  is one-to-one because it is increasing on its entire domain
Click for each step.
and, thus, has an inverse function. The steps:
f ( x)  x 2  2, x  0
1) replace (x) with y;
y  x 2  2, x  0
2) interchange x and y;
x  y 2  2, y  0
3) solve for y;
y  x2
4) replace y with -1
f 1 ( x)  x  2
5) prove that ( º-1)(x) = x and (-1 º)(x) = x.
Continued on next slide.
Solution (cont.)
FINDING THE INVERSE OF A FUNCTION WITH A
RESTRICTED DOMAIN
As a check, graphs of  and -1 are helpful.
f ( x)  x2  2, x  0
f 1 ( x)  x  2
Be Happy! Do Algebra!
Developed by Irina Safina.
North Harris LSCS