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Chapter 1
Functions
Overview
• Definition
• Domain and Range
• Rates of Change
• Composition
• Transformations
• Absolute Value
• Inverses
2
Chapter 1.1
What is a Function?
3
A Function
• A rule, or relation, that goes from x to y, or x to f(x)
• Must be unique – for each value of x, have only one f(x)
• Eliminates confusion – for each input there is a unique output
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Function or Not?
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Clearly a function
• Each value of the argument, p, has a unique partner h(p)
(Note, the variables needn’t be x and y!)
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Clearly a function
• Can also use vertical line test!
Any vertical line can only intersect the function curve once.
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Function or Not?
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• A and B are functions
• C is not. For x = 0, y can be ±3
• Again, vertical line test
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Another non-function relationship
• Input: the first name of each of you
• Output: the last name of each of you
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Why do we care?
• Many mathematical principles are confused if we apply them
to non-functions
• Unique results, for example, can be very important
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Evaluating a Function
• Consider the function f(x) = (x2 - 45)/3
• Evaluate f(x) for x = 9
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Solution
• Consider the function f(x) = (x2 - 45)/3
• Evaluate f(x) for x = 9
• Put 9 everywhere there is an x:
f(9) = (92 – 45)/3 = (81-45)/3 = 36/3 = 12
• f(9) = 12
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One to One Functions
• Each output value corresponds to only one input value
• Examples of functions that are not one-to-one:
f(x) = 5
f(x) = x2
f(x) = sin(x)
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A Test
• A one-to-one function meets the horizontal line test
• Here, f(0)= f(2) = 1; not one-to-one
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Example
• Is this a function? Is it one-to-one?
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Solution
• It is not a function
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Example
• Is this a function? Is it one-to-one?
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Solution
• It is a function
• It is not one-to-one
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• Is this a function? Is it one-to-one
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Solution
• This is a one-to-one function
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Summary
• A function assigns a unique value, f(x) to every input value, x
• A one-to-one function is a function that has a unique input x,
for every output f(x)
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Chapter 1.2
Domain and Range
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Overview
• Domain and range
• Piece-wise functions
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Domain and Range
• Domain is the values that go into the function: you live in your
domain
• Range is the values that come out of the function: you run
around in your range
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Domain and Range
• Specifying a domain may
– Exclude points for which a function is not defined
– Restrict the input values such that we have a function
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Example
• What are the domain and range of:
a. { (2, 5), (4, 8,), (1, 2)}
b. {(4, 5), (4, 6), (1, -5)}
Are they functions?
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Solution
• What are the domain and range of
a. { (2, 5), (4, 8), (1, 2)}
b. {(4, 5), (4, 6), (1, -5)}
a: Domain {2, 4, 1}, Range{5, 8, 2} yes a function
b: Domain {4, 1} Range {5, 6, -6} no, not a function;
for an input of 4 have two different outputs
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Notation
• Unfortunately, since there are so many applications of
mathematics, there are many notations used to represent
domain and range
• Use whatever notation you are comfortable with, however,
you should learn to understand the rest
• Whatever you do – don’t panic over notation! Use words, if
required. Many famous mathematicians, for centuries, have
done so
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Some Notation
• We can express range and domain in multiple ways
– Number line: open circle means point is not included
– Intervals: “(“ means point is not included
[1,3] ∪ (5, ∞)
– Inequalities: 1 ≤ 𝑥 ≤ 3, 𝑎𝑛𝑑 5 < 𝑥
• How would you express the following line in interval notation?
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-10 -0 10
Solution
• Notation: (-10, ∞)
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-10 -0 10
Yet one more…
• Set notation:
{x | 10 < x }
• Reads as: the set of values for x such that 10 is less than x
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Recap
• A function is a rule x → f(x)
– Each input, x, has a unique output, f(x)
– Vertical line test
• For a one to one function
– Each output, y, has a unique input, x
– Horizontal line test
• Domain: the allowable inputs to a function
• Range: the outputs of a function
• Notations: number line, intervals, set notation
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Example of Functions to Watch For
• Rational Functions: 4
5−𝑥
• Square Roots: 1 − 5𝑥
• Logarithms: log (1-x)
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Each of these functions has restricted domains
Comments
• Rational Functions: 4
5−𝑥
Domain cannot include x=5
• Square Roots: 1 − 5𝑥
Domain cannot include x > 1/5
• Logarithms: log (1-x)
Domain cannot include x 5
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Constant Function
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Not 1 to 1
Identity Function
• f(x) = x
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1 to 1
Absolute Value
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f(x) = |x|
Not 1 to 1
Quadratic: f(x) = x2
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Not 1 to 1
Cubic: f(x) = x3
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1 to 1
Reciprocal: f(x) = 1/x
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1 to 1
Square Root: f(x) = 𝑥
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1 to 1
Piecewise Functions
• Quite simply, functions that are defined in pieces
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Example: Graph
44
Note open and closed circles!
Example
• Determine if the following is a function, then graph and give
the range and domain of:
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Solution
This is a function.
Domain is x≠ 0
Range is y≠ ∞,−∞
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Find the range and domain
• f (x) = 7 − 𝑥
• 𝑓 𝑥 =2
1−𝑥
• 𝑓 𝑥 = 𝑐
• 𝑓 𝑥 = 𝑥
• 𝑓 𝑥 = 𝑥3
47
Solutions
• f (x) = 7 − 𝑥 :x ≤ 7, 𝑦 ≥ 0
• 𝑓 𝑥 =2
1−𝑥: 𝑥 ≠ 1, 𝑦 ≠ 0
• 𝑓 𝑥 = 𝑐 : x = all numbers, y = c
• 𝑓 𝑥 = 𝑥 : x = all numbers, y = all numbers
• 𝑓 𝑥 = 𝑥3 : x = all numbers, y = all numbers
48
Examples
Find the domain of
y = 3𝑥 + 9
y = 1
𝑡2 −6𝑡 −7
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Solution
Find the domain of
y = 3𝑥 + 9
If we are looking for y to be a real number, we need to
have 3x + 9 ≥ 0. Our domain is x ≥ -3
y = 1
𝑡2 −6𝑡 −7
We need to avoid dividing by zero, so our domain is
all real numbers, excluding 7 and -1
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Example
Find the domain of
𝑥+3
𝑥 −4
3 𝑥+3
𝑥 −4
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Solution
• For both of these expressions, we need x ≠4
• For the first, we also want the value under the square root
to be ≥ 0. This restricts x to be x ≤ -3 υ x > 4
𝑥+3
𝑥 −4
• For the second, there is no restriction that the value be ≥ 0, so
we have only x ≠4
3 𝑥+3
𝑥 −4
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Finding the Range
Find the range of
y = 𝑥+2
𝑥 −3
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Solution
Solve for x:
y = 𝑥+2
𝑥 −3
y(x-3) = x + 2
xy – x = 3y + 2
x = 3𝑦+2
𝑦 −1
We can see that for any y there is a x, excluding y = 1. However,
if y = 1 is a solution, that means 2 = -3, which is impossible
The range is all reals excluding y = 1
NOTE: This technique does not always work!54
If can graph, can find domain and range
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Some Functions You Will Need to Know
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y=|x| y=x2
y=x3y= 𝑥
y= 1 − 𝑥2y=1/x
Practice
• Try the exercises in section 1.2
• Bring “problem children” to class tomorrow
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Chapter 1.3
Rates of Change and Behavior of Graphs
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Overview
• Average Rate of Change
• Increasing and decreasing functions
• Local and absolute maxima and minima
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Rate of Change
• Change in output / change in input
• Change in range/ change in domain
• Change in y / change in x
• Change in f(x) / change in x
•∆𝑦
∆𝑥
• (y1-y2) / (x1-x2)
• ( f(x1)- f(x2)) / (x1-x1)
• What is another term for rate of change?
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• Rate of change is slope!
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Example
• Find the average rate of change of
f(x) = 3x2 + 4
on the interval (-3, 4)
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Solution
• Find the average rate of change of
f(x) = 3x2 + 4
on the interval (-3, 4)
• f(-3) = 27 + 4 = 31
• f(4) = 48 + 4 = 52
• f(3) – f(4) = 31-52 = -21
• -3 – (4) = -7
• Average rate of change is -21 / -7 = 3
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Example
• Find the average rate of change of
f(x) = x3 – x2 + 4
on the interval (0, a)
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Solution
• Find the average rate of change of
f(x) = x3 – x2 + 4
on the interval (0, a)
• f(0) =4
• f(a) = a3 – a2 + 4
• Rate of change is 𝑎3− 𝑎2
𝑎= a2 – a
65
Recap
• Average rate of change of a function f(x) on an interval [a,b]
𝑓 𝑎 − 𝑓(𝑏)
𝑎 − 𝑏
• For a line, is the same as slope: 𝑦1−𝑦2
𝑥1−𝑥2
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Find the Average Rate of Change
• p(x) = 3x + 4 on [2, 2+h]
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Solution
• p(x) = 3x + 4 on [2, 2+h]
p(2) = 6+4=10, p(2+h) = 6 + 4 + 3h = 10+ 3h
p(2) – p(2+h) = -3h
2 – (2+h) = -h
Average rate of change = 3
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Increasing and Decreasing Functions
69
• The function 𝑓(𝑥) = 𝑥3 − 12𝑥 is
increasing on (−∞, −2) ∪ (2, ∞) and
is decreasing on (−2, 2).
• Increasing means rate of
change > 0
• Decreasing means rate of
change< 0
Maxima and Minima
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Local maximum at b means f(b) ≥f(x)
for x in the interval (a, c) and a < b< c
Local minimum at b means f(b) ≤ f(x)
for x in the interval (a, c) and a < b< c
Local maximum at x = = - 2, y = 16
Local minimum at x = 2, y = -16
Local Min
Local Max
Formal Definitions
• A function f is an increasing function on an open interval if f (b) > f (a)
for any two input values a and b in the given interval where b > a
• A function f is a decreasing function on an open interval if f (b) < f (a)
for any two input values a and b in the given interval where b > a
• A function f has a local maximum at x = b if there exists an interval
(a, c) with a < b < c such that, for any x in the interval (a, c), f (x) ≤ f (b)
• Likewise, f has a local minimum at x = b if there exists an interval
(a, c) with a < b < c such that, for any x in the interval (a, c), f (x) ≥ f (b)
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Example
• Where is this function increasing or decreasing?
• Are there local minima, maxima?
72
Solution
• Increasing 1<x<3 and x>4
• Decreasing x<1 and 3<x<4
• Local Minima:
(1,-1), (4, ~0.6)
• Local Maxima
(3, 1)
73
Absolute Maximum and Minimum
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The absolute maximum of f at x = c is f (c) where f (c) ≥ f (x)
for all x in the domain of f .
The absolute minimum of f at x = d is f (d) where f (d) ≤ f (x)
for all x in the domain of f
Chapter 1.4
Composition of Functions
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Overview
• Combine functions using algebra
• Composition of functions
• Evaluating compositions
• Domain of composed functions
• Decomposing a composed function
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Arithmetic Operations
• (f + g)(x) = f(x) + g(x)
• (f - g)(x) = f(x) - g(x)
• (fg)(x) = f(x)g(x)
• (f / g)(x) = f(x) / g(x), for g(x) ≠ 0
• The domain is all inputs belonging to both f and g, with the
added provision that in the division g(x) ≠ 0
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Examples
If f(x) = x and g(x) = x - 1
• (f + g)(x) = f(x) + g(x) = 2x -1
• (f – g)(x) = f(x) – g(x) = 1
• (fg)(x) = f(x)g(x) = x2 – x
• (f/g)(x) =f(x)/g(x) = 𝑥
𝑥−1for x ≠ 1
We can also include constants:
• (f + 2g)(x) = f(x) + 2g(x) = 3x - 2, for example
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Recap: Combining Functions
• (f+g)(x) = f(x) + g(x)
• (f-g)(x) = f(x) g(x)
• (fg)(x) = f(x)g(x)
• (f/g)(x) = f(x)/g(x), g(x)≠ 0
• The domain is all inputs belonging to both f and g, with the
added provision that in the division g(x) ≠ 0
79
Composition of Functions
If f(x) = 3x and g(x) = x - 1
• f(g(x)) = 3(x-1)
We also write this as f g(x); use g(x) as the input to f
The domain of f g(x) is those inputs in the domain of g for
which g(x) is in the domain of f
Note: In general, f g(x) ≠ g f(x)
We say f composed with g or, simply, f of g
80
Examples
f(x) = x2 and g(x) = 2x + 2
Find f g(x) and g f(x)
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Solution
f(x) = x2 and g(x) = 2x + 2
Find f g(x) and g f(x)
f g(x) = (2x+2) 2: Domain is all numbers, range is y4
g f(x) = 2x2 + 2 : Domain is all numbers, range is y 2
82
Example
f(x) = x2 + 3 and g(x) = 𝑥
What is the domain of f(g(x))?
83
Solution
• f(x) = x2 + 3 and g(x) = 𝑥
• f(g(x)) = x +3, but the domain is not all real numbers.
• The domain of g is all positive reals,
hence the domain of f(g(x))is all numbers >3;
it is not the same as the domain of f(x)
84
More on Domain and Composition
85
𝑓 𝑥 =1
𝑥+2and 𝑔 𝑥 =
𝑥
𝑥−3
Continuing: Domain of f(g(x))
𝑓 𝑥 =1
𝑥+2and 𝑔 𝑥 =
𝑥
𝑥−3
3 is not in the domain of g(x), so it is not in the domain of f(g(x))
Additionally, we can’t have f(-2), so g(x) -2.
For g(x) = -2, we need -2 = 𝑥
𝑥−3or x = 2
So, the domain of f(g(x)) is all reals, but x 3 and x 2
86
What about g(f(x))?
𝑓 𝑥 =1
𝑥+2and 𝑔 𝑥 =
𝑥
𝑥−3
• To satisfy f, x 2
• To satisfy g, f(x) 3
for f(x) = 3, we have 3 = 1
𝑥+2, 3x + 6 = 1
Thus, x -5/3
• Domain is x 2, x -5/3
87
More Examples
Find f(g(x)) and g(f(x)) and the domain of each
• f(x) = 𝑥 − 2, g(x) = 1
3−𝑥
• f(x) = 𝑥−3
2𝑥, g(x) =
3
𝑥
88
Solution
Find f(g(x)) and g(f(x)) and the domain of each
• f(x) = 𝑥 − 2, g(x) = 1
3−𝑥
f(g(x)) = 1
3−𝑥− 2 : x≠ 3 and 1/(3-x) ≥ 2; the second gives
3-x ≤ ½, or x ≥ 5/2
g(f(x))= 1
3− 𝑥−2; need x≥ 2 and x≠ 11
89
Solution
Find f(g(x)) and g(f(x)) and the domain of each
• f(x) = 𝑥−3
2𝑥, g(x) =
3
𝑥
f(g(x)) = [3/x – 3] / [6/x] = (3 – 3x)/6 = (1 – x)/2
can’t have x = 0 because of the definition of g.
Range is x≠ 0
g(f(x)) = 3/[(x-3)/2x] = 6x/(x-3)
can’t have x = 0 because of the definition of x, also
cannot have x=3
Range is (−∞, 0) ∪ (0,3) ∪ (3,∞)
90
Example
• A store offers customers a 30% discount on the price x of
selected items. Then, the store takes off an additional 15% at
the cash register.
Write a price function P(x) that computes the final price of the
item in terms of the original price x.
91
Solution
• A store offers customers a 30% discount on the price x of
selected items. Then, the store takes off an additional 15% at
the cash register.
Write a price function P(x) that computes the final price of the
item in terms of the original price x.
Final price = register discount (price discount (item))
FP(x) = 0.85 ( 0.7 (x))
(0.85)(0.7) = 0.595
FP(x) = 0.595 x
92
Summary
• When composing functions, f(g(x)), the domain is only
those values where g(x) is defined and f(y), where y = g(x),
is defined
BE CAREFUL
• f(a+b) f(a) + f(b)
• f(ab) f(a)(f(b)
• f(a/b) f(a)/f(b)
• f(g(x)) g(f(x))
93
Chapter 1.5
Transformation of Functions
94
Overview
• Vertical and horizontal shift
• Even and odd functions
• Compression and stretching
• Transformation combinations
95
Vertical Shift
96
• Move the graph upwards by k units
• G(x) = f(x)+k
• Here we have the vertical shift by 𝑘 = 1 of the cube root function 𝑓(𝑥) = 3 𝑥.
Horizontal Shift
97
• G(x) = f(x+k) shifts the function k units to the left
• Horizontal shift of the function 𝑓(𝑥) = 3 𝑥.
• Note that ℎ = + 1 shifts the graph to the left, that is, towards negative values of x.
How to do a shift?
You can either:
1. Use a table, as when you graph a function
2. Move the function on the graph itself
For example, f(x) = 4x
98
x f(x) f(x+3) f(x)+3
-2 -8 4 -5
-1 -4 8 -1
0 0 12 3
1 4 16 7
2 8 20 11
-10
-5
0
5
10
15
20
25
-4 -2 0 2 4
f(x)
f(x+3)
f(x)+3
Example
• For f(x) = x2 + 3, graph f(x), f(x+2), f(x)+2
99
Solution
• For f(x) = x2 + 3, graph f(x), f(x+2), f(x)+2
100
x f(x) f(x+2) f(x)+2
-2 7 0 9
-1 4 1 6
0 3 4 5
1 4 9 6
2 7 16 9
Combining Horizontal and Vertical Shifts
• Suppose f(x) = |x|, h(x) = f(x+1) + 3. Graph h(x)
101
Solution
102
x f(x) f(x+1) h(x)
-3 3 2 5
-2 2 1 4
-1 1 0 3
0 0 1 4
1 1 2 5
2 2 3 6
3 3 4 7
Shift Summary
• g(x) = f(x)+k, move k units up in y
• g(x) = f(x)-k, move k units down in y
• g(x) = f(x+k), move k units to the left in x
• g(x) = f(x-k), move k units to the right in x
• Question: Why does positive in k make the graph go negative
in x and positive in y?
103
Solution
• g(x) = f(x)+k, move k units up in y
• g(x) = f(x)-k, move k units down in y
• g(x) = f(x+k), move k units to the left in x
• g(x) = f(x-k), move k units to the right in x
• Question: Why does positive in k make the graph go negative
in x and positive in y?
• If we rewrite: g(x)-k = f(x), it goes up in y, and the convention
is the same, etc.
104
Reflections
105
• Given a function f (x), a new function g(x) = − f (x) is a vertical reflection of
the function f (x), sometimes called a reflection about the x-axis.
• Given a function f (x), a new function g(x) = f ( − x) is a horizontal reflection
of the function f (x), sometimes called a reflection about the y-axis.
Example
• f(x) = x2; graph g(x) = -f(x), h(x) = f(-x)
106
Solution
• f(x) = x2; graph g(x) = -f(x), h(x) = f(-x)
Note f(x) = h(x)!
107
x f(x) g(x) h(x)
-3 9 -9 9
-2 4 -4 4
-1 1 -1 1
0 0 0 0
1 1 -1 1
2 4 -4 4
3 9 -9 9
Example
Sketch the following, starting with y = x2 :
y = x2 + 3
y = (x-4) 2 + 3
y = -(x-4) 2 - 3
108
Solution
Sketch the following, starting with y = x2 :
y = x2 + 3
y = (x-4) 2 + 3
y = -(x-4) 2 - 3
109
Example
• Use the graph y = |x| to graph y = - |x - 5| - 4
110
Solution
• Use the graph y = |x| to graph y = - |x - 5| - 4
111
Vertical Scaling (Compression and Stretching)
• If f is a function, and c a constant >0, replacing y by cy gives
y = 1/c f(x)
If c > 1, the graph is compressed by a factor of c
If c < 1, the graph is stretched by a factor of c
112
Vertical Scaling Example
• Start with y = 2x – x2, 0 ≤ x ≤ 2
• Now, consider 2y = 2x – x2 or y = ½(2x - x2)
• Now, consider 1/2y = 2x – x2
113
Horizontal Scaling
• y = f(x) goes to y = f(cx)
• Compare y = x2 with y = (3x)2 and y = (x/3) 2
114
Solution
115
x x*x (3x)*(3x) (x/3)*(x/3)
-3 9 81 1
-2 4 36 0.4444444
-1 1 9 0.1111111
0 0 0 0
1 1 9 0.1111111
2 4 36 0.4444444
3 9 81 1
Compare y = x2 with y = (3x)2 and y = (x/3) 2
0
10
20
30
40
-4 -2 0 2 4
x*x (3x)*(3x) (x/3)*(x/3)
Transformation Summary
• Let (a,b) be on the graph of y = f(x) and c a positive constant
116
Transformation Change New Equation New Point
Shift up c units y y – c y – c = f(x) (a, b+c)
Shift down c units y y + c y + c = f(x) (a, b-c)
Shift right c units x x – c y = f(x-c) (a+c, b)
Shift left c units y x + c y = f(x+c) (a-c, b)
Reflection in x-axis y -y y = -f(x) (a, -b)
Reflection in y-axis x -x y = f(-x) (-a, b)
Vertical Scale Change
Horizontal Scale Change
y cy
x cx
y = f(x)/c
y =f(cx)
(a, b/c)
(a/c, b)
Be careful with reflection and sign!
117
• Graph y = −𝑥 + 2
– So far we have looked at y = f(xc)
– What is different when we have f(-xc)
• Graph y = −𝑥 + 2
• Start with y = 𝑥 and shift to the left 2 units, y = 𝑥 + 2
• Replace x by –x and reflect the graph in the x axis
118
Be careful with reflection and sign!
• It might help to think of working from the
“parentheses” inwards
• Graph 1
1−𝑥
• If we graph 1/x and shift to the left by 1, then reflect, we will
have the wrong answer
• Instead, graph 1/x, then reflect, then move to the right by 1
119
Using Transformation
Graph y = 4(x-5)2 -3
• Start with y = x2
• Vertically stretch y = 4x2
• Replace x by x-5
• Replace y by y+3
There is no unique procedure/order!
You can always check by plugging in a point
120
How do we find a transformation from equation?
• Suppose we had
y = 1/3 f(-x/2 + 4) + 5
121
Solution
Finding a transformation from an equation
• Suppose we had
y = 1/3 f(-x/2 + 4) + 5
• If we start with y = f(x)
– Horizontal scale change of ½ -- graph is stretched by a
factor of two
– Replace x by -x, reflection on y axis
– Substitute x-8 for x, creating a shift to the right
– Substitute 3y for y, a vertical scale change (compression)
– Now shift the graph up by 5 units
122
Solution
Graph y = 4(x-5)2 -3
123
Graph using Transformation
• y = 1/(5-x)
• y = 𝑥 − 3 + 1
• y = - 1 − (𝑥 − 2)2
124
y = 1/(5-x)
125
Blue is basic 1x
Red is 1/(-x)
y = 𝑥 − 3 + 1
126
Blue: Initial
Red: H shift 3
Green: V shift 1
y = - 1 − (𝑥 − 2)2
127
Blue: 1 − 𝑥2
Red: - 1 − 𝑥2
Green: Final
And a Point?
If (a,b) was on the graph of y = f(x), what is that point on
y = 1/3 f(-x/2 + 4) + 5
• (a,b) (2a, b) (-2a, b) (-2a – 8,b) (-2a – 8, b/3)
(-2a+8, b/3 + 5)
128
Even and Odd Functions
• Even functions: f(-x) = f(x)
• Odd functions: f(-x) = -f(x)
• For a function to be even, the rule must hold for each value in
the domain
• Same with an odd function
• There can be no points of exceptions; a function is even, odd,
or neither
• The only function that is both even and odd is f(x)=0
129
Even, Odd or Neither?
• f(x) = 1 − 𝑥2
2+ 𝑥2
• g(x) = 𝑥 − 𝑥3
2𝑥+ 𝑥3
• h(x) = (x2 + x) 2
• f(x) = x |x| / 𝑥 − 4
1 if x > 0
• g(x) = 0 if x = 0
-1 if x < 0 130
Solution
• f(x) = 1 − 𝑥2
2+ 𝑥2even
• g(x) = 𝑥 − 𝑥3
2𝑥+ 𝑥3even
• h(x) = (x2 + x) 2 neither
• f(x) = x |x| / 𝑥 − 4 odd
1 if x > 0
• g(x) = 0 if x = 0 odd
-1 if x < 0 131
Recap
• We can create graphs of many functions by graphing
canonical functions and then modifying them
• Some functions you should know:
y = x2
y = 𝑥
y = 1 − 𝑥2
y = 𝑥3
y = |x|
y = 1/x
132
Modification Techniques
• Horizontal shift: x becomes x ± a
• Vertical shift: y becomes y ± a
• Compression/expansion: x or y becomes cx or cy
• Reflection in x or y axis
133
General Rules
• When combining vertical transformations written in the form
a f (x) + k, first stretch by a and then shift by k
• When combining horizontal transformations written in the form
f (bx + h), first shift by h and then stretch by b
• When combining horizontal transformations written in the form
f (b(x + h)), first horizontally stretch by b and then horizontally
shift by h
• Horizontal and vertical transformations are independent. It
does not matter whether horizontal or vertical transformations
are performed first.
134
Key things to watch for:
1. Watch out when going from –x to –x – a or –x + a
Example: creating y = 1
−𝑥+𝑎
– Either reflect x to get –x before shifting by a or reflect x – a
to get –x + a
2. Be careful of f(ax – c)
May be best to work with f(a[x -𝑐
𝑎] )
135
General Rule, Again
• When combining horizontal transformations written in the form
• f (bx + h), first shift by h and then stretch by b
• When combining horizontal transformations written in the form
f (b(x + h)), first horizontally stretch by b and then horizontally
shift by h
• Think about the parentheses! Does the transform include the
whole argument? If so, do that first!
• When in doubt, make a table of key points!
136
Chapter 1.6
Absolute Value
137
Overview
• Graph an absolute value
• Solve an absolute value equality
• Solve an absolute value inequality
138
Absolute Value as a Piecewise Function
• The absolute value function can be defined as a
piecewise function
• f (x) = |x| = x if x 0
−x if x < 0
• We can also think of |a| as the distance of a from zero;
distance is always a positive quantity
139
Examples
• |5| = 5, the distance of 5 from zero
• What values satisfy |x-4| = 3?
140
Solution
• What values satisfy |x-4| = 3?
• We need the distance between x and 4 to be 3
x can be 3 greater than 4 or 3 less than 4
x = 7 or x = 1
141
Uses
• Tolerances:
The diameter of a rod must be 2 inches 10%
• 10% of 2 inches is 0.2 inches
• The diameter can be as small as 1.8 inches or as large
as 2.2 inches
142
Graphing Absolute Values
• Most important is where the direction changes
143
f(x) = 2|x-3|+4
144
In Pieces
• Write the equation for the following:
145
Solution
• Origin is shifted right by 3 and down by 2, f(x) f(x-3)-2
• Function is also stretched, vertically, by 2, or f(x) 2f(x)
• Function is f(x) = 2|x-3| - 2; plug in points to verify!
146
Solution Cont.
• Function is f(x) = 2|x-3| - 2 ; plug in points to verify!
x = 0, f(0) = 6-2=4
x = 3, f(3) = 0-2= -2
x=2, f(2) = 2(1)-2 = 0; etc.
147
Figuring out the “stretch”
• F(x) = a|x-3| - 2, need to find a
• Plug in a point, e.g., x = 0
f(0)=4 = a|0-3|-2 = 3a – 2
3a = 4+2, a = 2
148
Note:
• F(x)=a|x-v|+h
– Always intersects the y axis
– May intersect the x axis
149
• The absolute value graph can intersect the x axis at
0, 1, or 2 points
150
How to solve?
• We can always break the equality into two pieces
K = |f(x)| becomes:
K = -f(x) AND K = f(x)
151
Some More Equalities
• Find x:
|3-x| = 10
|x-1|/3= 4
4|3-x| = 8
152
Solutions
• Find x:
|3-x| = 10 ; 3-x = 10, x = -7
x-3 = 10, x = 13
|x-1|/3= 4; (x-1)/3 = 4, x-1=12, x = 13
(1-x)/3=4, 1-x = 12, x = -11
4|3-x| = 8 ; 4(3-x) = 8, (3-x)=2, x = 1
4(x-3) = 8, (x-3)= 2, x = 5
153
Intersecting the x axis
• Where do these equalities intersect the x axis?
|3-x| = 10
|x-1|/3= 4
4|3-x| = 8
154
Solution
• Where do these inequalities intersect the x axis?
• Along the x axis, y = 0; just like solving the equations for f(x)=0
|3-x| = 10
0= 10 – (3-x), x = -7
0 = 10-(x-3), x = 13
|x-1|/3= 4
0 = - 4 + (x-1)/3, 12 = x-1, x = 13
0 = -4 – (x-1)/3, -12 = x-1, x = -11
4|3-x| = 8
0 = 8 -4(3-x), x = 1
0 = 8 +4(3-x), x = 5
155
Do Abs Val Eqs always have two answers?
156
Do Abs Val Eqs always have two answers?
• No
• Remember, the absolute value can intersect the x axis zero,
one or two times
• There may be no solution!
157
Example
• Where does f(x) = 4|x-2| + 2 intersect the x axis?
158
Solution
• Where does f(x) = 4|x-2| + 2 intersect the x axis?
• 0 = 4|x-2|+2
-2 = 4|x-2|
-1/2 = |x-2| but |x-2| can never be -1/2; it can’t be < 0
• No solution
159
Solving an inequality
• Find the values of x for which
F(x) = |x-3|+4 > 2
160
Solution
• Find the values of x for which
F(x) = |x-3|+4 > 2
• 2< 4+|x-3|
-2 < |x-3|
• |x-3| is always > -2, so holds for all x
161
More examples
• f(x) = −|4x − 5|/2 + 3, where is f(x)<0?
162
Solutions
• f(x) = −|4x − 5|/2 + 3, where is f(x)<0?
-3< -|4x-5|
Multiply by -1 and change the sign: 3>|4x-5|
Now have two problems: 3 > 4x-5 and 3>5-4x
This gives x<2 and x>-1/2
or (-1/2, 2)
Can always check by plugging in a point. For example:
If x = -1, we have -3 < - 9, doesn’t work
If x = 0 we have -3 < -5, works
If x = 3, we have -3 < -|12-5| = -7, doesn’t work
163
More examples
• 3|x-5|<15
• |2x-4|<5
• 3|x+1| - 4 > -1
164
Solutions
• 3|x-5|<15
|x-5|<5: x-5<5 giving x<10 and 5-x<5, giving –x < 0, x>0
Need 0<x<10
• |2x-4|<5
2x-4<5 or 2x<9, x<9/2 and 4-2x > -5, or 2x<-1, x<-1/2
Need -1/2<x<9/2
• 3|x+1| - 4 > -1
3|x+1|>3, |x+1|>1, x+1 > 1, x > 0 and x+1 <-1 or x<-2
Need x < -2 OR x > 0
165
When do we have what answer?
• When do we have a<x<b and
when do we have a < x or x < b?
• Some basic rules:
– If |f(x)| <0, there is no solution
– If |f(x)| < a, where a > 0, is an interval solution (c, d)
– If |f(x)| > a, where a > 0, is a union solution x< a AND x>b
– If |f(x)|>0, any x is a solution
166
Recap: Solving Equations and Inequalities
• |f(x)| = a
– Want the absolute value of f(x) to be a
– Set f(x) = a and f(x) = -a and solve for x
– If a<0, is no solution
• |f(x)| < a
– Want absolute value of f(x) to be <a
– Set f(x) <a and f(x)>-a and solve for x
– Can be no solution or one of the form p < x < q,
p and q numbers
• |f(x)|>a: set f(x)>a and f(x)<-a and solve for x
solution may be of the form x<p OR x>q
167
Chapter 1.7
Inverse Functions
168
Overview
• Verify inverse functions.
• Determine the domain and range of an inverse function, and
restrict the domain of a function to make it one-to-one.
• Find or evaluate the inverse of a function.
• Use the graph of a one-to-one function to graph its inverse
function on the same axes.
169
What is an Inverse Function?
• The inputs become the outputs and the outputs the inputs
• If you do the function, then the inverse, should get back where
you started…
• From your text:
170
Note
• For a function to have an inverse, it must be one-to-one
• Why?
171
Why?
• If a function is one-to-one, each output comes from a unique
input value
• Since the outputs become the inputs of the inverse function,
this is required for the inverse to be a function
172
Notation
• The inverse of f(x) is f-1(x)
• f-1(x) does not mean 1/ f-1(x)
173
Range and Domain
• If a function, f is one-to-one, then the inverse of f(x),
f -1(x) is a function such that f -1(f(x)) = x for x in the domain
of f, and f(f -1 (y))= y for y in the domain of f -1
• Note: the range of the inverse is equal to the domain of the
original function; the domain of the inverse is the range of the
original function
174
Verifying Functions are Inverses
• Need:
f-1(f(x)) = x
f(f-1(x)) = x
For example, the inverse of f(x) = 2x is f-1(x) = x/2
f(f-1(x)) = 2(f-1(x)) = 2(x/2) = x, and
f-1(f(x)) = f-1(2x)= (2x)/2 = x
175
Verify the Following Inverses
• f(x) = 1/(x+3), f-1(x) = 1/x - 3
176
Solution
• f(x) = 1/(x+3), f-1(x) = 1/x – 3
f(f-1(x)) = 1/[(1/x – 3) + 3]
f(f-1(x)) = 1/[1/x] = x
f-1(f(x)) = 1/[1/(x+3)] – 3
f-1(f(x)) = x+3 – 3 = x
177
Example
• Is g(x) = ½ x the inverse of f(x) = x2?
178
Solution
• Is g(x) = ½ x the inverse of f(x) = x2?
g(f(x)) = ½ (x2) ≠ 𝑥
179
Example
• Show the inverse of x4 is 4 𝑥
180
Solution
• Show the inverse of x4 is 4 𝑥
4𝑥4 = x, are inverses
181
How do we find the inverse?
• Given y = f(x), solve for x = f-1(y)
182
Example
• f(x) = 𝑥 . Find f -1 (y), its domain and range
183
Solution
• f(x) = 𝑥 . Find f -1 (y), its domain and range
• y = 𝑥; solve for x
• Square both sides: y2 = x
• So, f -1 (y) = y2
184
Solution: Domain and Range
• x = y2, so f -1 (y) = y2
• What is the domain of f -1? It must be the range of f(x),
or y 0; additionally, we have to have y 0 for f -1 (y) to
be one to one
Remember, the range of the inverse is equal to the domain of
the original function; the domain of the inverse is the range of the
original
• The domain of the original function, f(x) = 𝑥, is x 0, so this
is the range of the inverse, f -1 (y)
185
Example
• Find the inverse of y = f(x) = 3
2+𝑥and its domain and range
186
Solution
• Find the inverse of y = f(x) = 3
2+𝑥and its domain and range
• Set f(x) = y = 3/(2+x) and solve for x
• (2+x)y = 3
• 2+x = 3/y
• x = f-1(y) = (3/y) -2 = (3 – 2y)/y
187
Solution, Domain and Range
• f(x) = 3
2+𝑥
• f-1(y) = 3−2𝑦
𝑦
• Domain of f(x) is x 2, domain of f-1(y) is y 0
• Therefore, the range of f (x) is x 0 and
the range of f-1(y) is y 2
• Is there any way f(x) can be zero? 0 = 3/(2+x), 0 = 3: NO!!!
• How about f-1(y) = 2? 2 = (3-2y)/y or 2y = 3-2y, 0 = 3 NO!!!
• Can also test f(f-1(y))=y
f(f-1(y))=3/(2+(3-2y)/y) = 3/([2y + 3 – 2y]/y) = 3y/3 = y; Works!
188
Find the inverse, domain and range of f(x)
• f(x) = 2𝑥+3
𝑥 −1
189
Solution
• f(x) = 2𝑥+3
𝑥 −1
• y = (2x + 3)/(x – 1)
• yx – y = 2x + 3
• x(y – 2) = 3 + y
• x = (3 + y)/(y-2)
• Range and domain: We can’t have y = 2 in the domain of the
inverse. Is it in the range of f?
2 = (2x+3)/(x-1), 2x -2 = 2x + 3. But -2 is never 3!
• Can 1 be in the range of the inverse?
1 = (3+x)/(x-2), x – 2 = 3 + x, again, it doesn’t work
• Domain of is x1, range is y 2 190
Example
• f(x) = x2 – 2x – 3; find the inverse function
191
Solution
• f(x) = x2 – 2x – 3; find the inverse function
• Does this function have an inverse?
• In order for it to have an inverse, we have to restrict it to be
one-to-one.
• Let’s go forward and see what happens
192
Solution
• f(x) = x2 – 2x – 3 = y; find the inverse function
• How do we solve?
Try completing the square:
(x-1)2 – 3 -1 = y
(x-1)2 = y-4, x-1 = ± 𝑦 − 4
We have f-1(y) = x = 1 ± 𝑦 − 4 as our inverse
Need it to be a function! This is not! Why?
193
Solution
• We needed our original function f(x) = y to be one-to-one
• y = x2 – 2x – 3; for x to be one-to-one, we can restrict x to
domain x < 1 or x > 1, not both
f-1(y) = x = 1 ± 𝑦 − 4
We choose the – sign if we chose x < 1 and vice versa
• f -1 (y) = 1 + 𝑦 + 4 or f −1 (y)1 + 𝑦 + 4
with the sign depending on our original definition of f(x)
194
Symmetry around y = x
• Points P and Q are symmetric around y = x if the segment PQ
is perpendicular to y = x and P and Q are equidistant from the
line y = x
• The graph of f −1(x) is the graph of f (x) reflected about the
diagonal line y = x, which we will call the identity line
195
Example
• Show (2, 5) and (5, 2) are symmetric about y = x
196
Solution
• Show (2, 5) and (5, 2) are symmetric about y = x
• Find the slope of a line through the points
(2-5)/(5-2) = -1, so is perpendicular
• The midpoint of the line segment connecting the two points is
3.5, 3.5, which is on the line y = x
197
Finding and Graphing an Inverse Function
• Let f(x) = 1
𝑥−3find and graph f -1 (y)
198
Graphing Inverse f(x) = 1
𝑥−3
199
Solution
• y = 1/(x-3)
• x-3 = 1/y
• x = 3 + 1/y = f -1 (y)
• Domain is y ≠ 0, range is y ≠ 3
200
Summary
• Definition: function, one to one
• Domain and Range: inputs and outputs
• Rates of Change: change in y / change in x, slope
• Composition: argument of a function is a function
• Transformations: vertical, horizontal, stretching
• Absolute Value: equalities and inequalities
• Inverses: f(f-1(x)) =x, domain, range; must be one to one funct
201
