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Functions are the fundamental concept in calculus. We discuss four ways to represent a function and different classes of mathematical expressions for functions.
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Section 1.1–1.2Functions
V63.0121.006/016, Calculus I
New York University
May 17, 2010
Announcements
I Get your WebAssign accounts and do the Intro assignment. ClassKey: nyu 0127 7953
I Office Hours: TBD
. . . . . .
. . . . . .
Announcements
I Get your WebAssignaccounts and do the Introassignment. Class Key:nyu 0127 7953
I Office Hours: TBD
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 2 / 54
. . . . . .
Objectives: Functions and their Representations
I Understand the definitionof function.
I Work with functionsrepresented in differentways
I Work with functionsdefined piecewise overseveral intervals.
I Understand and apply thedefinition of increasing anddecreasing function.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 3 / 54
. . . . . .
Objectives: A Catalog of Essential Functions
I Identify different classes ofalgebraic functions,including polynomial(linear, quadratic, cubic,etc.), polynomial(especially linear,quadratic, and cubic),rational, power,trigonometric, andexponential functions.
I Understand the effect ofalgebraic transformationson the graph of a function.
I Understand and computethe composition of twofunctions.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 4 / 54
. . . . . .
What is a function?
DefinitionA function f is a relation which assigns to to every element x in a set Da single element f(x) in a set E.
I The set D is called the domain of f.I The set E is called the target of f.I The set { y | y = f(x) for some x } is called the range of f.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 5 / 54
. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 6 / 54
. . . . . .
The Modeling Process
...Real-worldProblems
..Mathematical
Model
..MathematicalConclusions
..Real-worldPredictions
.model.solve
.interpret
.test
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 7 / 54
. . . . . .
Plato's Cave
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 8 / 54
. . . . . .
The Modeling Process
...Real-worldProblems
..Mathematical
Model
..MathematicalConclusions
..Real-worldPredictions
.model.solve
.interpret
.test
.Shadows .Forms
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 9 / 54
. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 10 / 54
. . . . . .
Functions expressed by formulas
Any expression in a single variable x defines a function. In this case,the domain is understood to be the largest set of x which aftersubstitution, give a real number.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 11 / 54
. . . . . .
Formula function example
Example
Let f(x) =x+ 1x− 2
. Find the domain and range of f.
SolutionThe denominator is zero when x = 2, so the domain is all real numbersexcept 2. As for the range, we can solve
y =x+ 2x− 1
=⇒ x =2y+ 1y− 1
So as long as y ̸= 1, there is an x associated to y. Therefore
domain(f) = { x | x ̸= 2 }range(f) = { y | y ̸= 1 }
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 12 / 54
. . . . . .
Formula function example
Example
Let f(x) =x+ 1x− 2
. Find the domain and range of f.
SolutionThe denominator is zero when x = 2, so the domain is all real numbersexcept 2.
As for the range, we can solve
y =x+ 2x− 1
=⇒ x =2y+ 1y− 1
So as long as y ̸= 1, there is an x associated to y. Therefore
domain(f) = { x | x ̸= 2 }range(f) = { y | y ̸= 1 }
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 12 / 54
. . . . . .
Formula function example
Example
Let f(x) =x+ 1x− 2
. Find the domain and range of f.
SolutionThe denominator is zero when x = 2, so the domain is all real numbersexcept 2. As for the range, we can solve
y =x+ 2x− 1
=⇒ x =2y+ 1y− 1
So as long as y ̸= 1, there is an x associated to y.
Therefore
domain(f) = { x | x ̸= 2 }range(f) = { y | y ̸= 1 }
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 12 / 54
. . . . . .
Formula function example
Example
Let f(x) =x+ 1x− 2
. Find the domain and range of f.
SolutionThe denominator is zero when x = 2, so the domain is all real numbersexcept 2. As for the range, we can solve
y =x+ 2x− 1
=⇒ x =2y+ 1y− 1
So as long as y ̸= 1, there is an x associated to y. Therefore
domain(f) = { x | x ̸= 2 }range(f) = { y | y ̸= 1 }
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 12 / 54
. . . . . .
No-no's for expressions
I Cannot have zero in the denominator of an expressionI Cannot have a negative number under an even root (e.g., square
root)I Cannot have the logarithm of a negative number
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 13 / 54
. . . . . .
Piecewise-defined functions
Example
Let
f(x) =
{x2 0 ≤ x ≤ 1;3− x 1 < x ≤ 2.
Find the domain and range of f and graph the function.
SolutionThe domain is [0,2]. The range is [0,2). The graph is piecewise.
...0
..1
..2
..1
..2
.
.
.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 14 / 54
. . . . . .
Piecewise-defined functions
Example
Let
f(x) =
{x2 0 ≤ x ≤ 1;3− x 1 < x ≤ 2.
Find the domain and range of f and graph the function.
SolutionThe domain is [0,2]. The range is [0,2). The graph is piecewise.
...0
..1
..2
..1
..2
.
.
.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 14 / 54
. . . . . .
Functions described numerically
We can just describe a function by a table of values, or a diagram.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 15 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,5,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 16 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,5,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 16 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,5,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 16 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 43 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 17 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 43 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 17 / 54
. . . . . .
Example
Is this a function? If so, what is the range?
x f(x)1 42 43 6
.
. .
..1
..2
..3
. .4
. .5
. .6
Yes, the range is {4,6}.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 17 / 54
. . . . . .
Example
How about this one?
x f(x)1 41 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
No, that one’s not “deterministic.”
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 18 / 54
. . . . . .
Example
How about this one?
x f(x)1 41 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
No, that one’s not “deterministic.”
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 18 / 54
. . . . . .
Example
How about this one?
x f(x)1 41 53 6
.
. .
..1
..2
..3
. .4
. .5
. .6
No, that one’s not “deterministic.”
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 18 / 54
An ideal function
. . . . . .
. . . . . .
Why numerical functions matter
In science, functions are often defined by data. Or, we observe dataand assume that it’s close to some nice continuous function.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 20 / 54
. . . . . .
Numerical Function Example
Here is the temperature in Boise, Idaho measured in 15-minuteintervals over the period August 22–29, 2008.
...8/22
..8/23
..8/24
..8/25
..8/26
..8/27
..8/28
..8/29
..10
..20
..30
..40
..50
..60
..70
..80
..90
..100
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 21 / 54
. . . . . .
Functions described graphically
Sometimes all we have is the “picture” of a function, by which wemean, its graph.
.
.
The one on the right is a relation but not a function.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 22 / 54
. . . . . .
Functions described graphically
Sometimes all we have is the “picture” of a function, by which wemean, its graph.
.
.
The one on the right is a relation but not a function.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 22 / 54
. . . . . .
Functions described verbally
Oftentimes our functions come out of nature and have verbaldescriptions:
I The temperature T(t) in this room at time t.I The elevation h(θ) of the point on the equator at longitude θ.I The utility u(x) I derive by consuming x burritos.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 23 / 54
. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 24 / 54
. . . . . .
Monotonicity
Example
Let P(x) be the probability that my income was at least $x last year.What might a graph of P(x) look like?
.
..1
..0.5
..$0
..$52,115
..$100K
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 25 / 54
. . . . . .
Monotonicity
Example
Let P(x) be the probability that my income was at least $x last year.What might a graph of P(x) look like?
.
..1
..0.5
..$0
..$52,115
..$100K
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 25 / 54
. . . . . .
Monotonicity
Definition
I A function f is decreasing if f(x1) > f(x2) whenever x1 < x2 forany two points x1 and x2 in the domain of f.
I A function f is increasing if f(x1) < f(x2) whenever x1 < x2 for anytwo points x1 and x2 in the domain of f.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 26 / 54
. . . . . .
Examples
Example
Going back to the burrito function, would you call it increasing?
Example
Obviously, the temperature in Boise is neither increasing nordecreasing.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 27 / 54
. . . . . .
Examples
Example
Going back to the burrito function, would you call it increasing?
Example
Obviously, the temperature in Boise is neither increasing nordecreasing.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 27 / 54
. . . . . .
Symmetry
Example
Let I(x) be the intensity of light x distance from a point.
Example
Let F(x) be the gravitational force at a point x distance from a blackhole.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 28 / 54
. . . . . .
Possible Intensity Graph
..x
.y = I(x)
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 29 / 54
. . . . . .
Possible Gravity Graph
..x
.y = F(x)
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 30 / 54
. . . . . .
Definitions
Definition
I A function f is called even if f(−x) = f(x) for all x in the domain of f.I A function f is called odd if f(−x) = −f(x) for all x in the domain of
f.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 31 / 54
. . . . . .
Examples
I Even: constants, even powers, cosineI Odd: odd powers, sine, tangentI Neither: exp, log
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 32 / 54
. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 33 / 54
. . . . . .
Classes of Functions
I linear functions, defined by slope an intercept, point and point, orpoint and slope.
I quadratic functions, cubic functions, power functions, polynomialsI rational functionsI trigonometric functionsI exponential/logarithmic functions
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 34 / 54
. . . . . .
Linear functions
Linear functions have a constant rate of growth and are of the form
f(x) = mx+ b.
Example
In New York City taxis cost $2.50 to get in and $0.40 per 1/5 mile. Writethe fare f(x) as a function of distance x traveled.
AnswerIf x is in miles and f(x) in dollars,
f(x) = 2.5+ 2x
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 35 / 54
. . . . . .
Linear functions
Linear functions have a constant rate of growth and are of the form
f(x) = mx+ b.
Example
In New York City taxis cost $2.50 to get in and $0.40 per 1/5 mile. Writethe fare f(x) as a function of distance x traveled.
AnswerIf x is in miles and f(x) in dollars,
f(x) = 2.5+ 2x
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 35 / 54
. . . . . .
Linear functions
Linear functions have a constant rate of growth and are of the form
f(x) = mx+ b.
Example
In New York City taxis cost $2.50 to get in and $0.40 per 1/5 mile. Writethe fare f(x) as a function of distance x traveled.
AnswerIf x is in miles and f(x) in dollars,
f(x) = 2.5+ 2x
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 35 / 54
. . . . . .
Example
Biologists have noticed that the chirping rate of crickets of a certainspecies is related to temperature, and the relationship appears to bevery nearly linear. A cricket produces 113 chirps per minute at 70 ◦Fand 173 chirps per minute at 80 ◦F.(a) Write a linear equation that models the temperature T as a function
of the number of chirps per minute N.(b) What is the slope of the graph? What does it represent?(c) If the crickets are chirping at 150 chirps per minute, estimate the
temperature.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 36 / 54
. . . . . .
Solution
I The point-slope form of the equation for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equation
y− y0 = m(x− x0)
I The slope of our line is80− 70
173− 113=
1060
=16
I So an equation for T and N is
T− 70 =16(N− 113) =⇒ T =
16N− 113
6+ 70
I If N = 150, then T =376
+ 70 = 7616◦F
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 37 / 54
. . . . . .
Solution
I The point-slope form of the equation for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equation
y− y0 = m(x− x0)
I The slope of our line is80− 70
173− 113=
1060
=16
I So an equation for T and N is
T− 70 =16(N− 113) =⇒ T =
16N− 113
6+ 70
I If N = 150, then T =376
+ 70 = 7616◦F
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 37 / 54
. . . . . .
Solution
I The point-slope form of the equation for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equation
y− y0 = m(x− x0)
I The slope of our line is80− 70
173− 113=
1060
=16
I So an equation for T and N is
T− 70 =16(N− 113) =⇒ T =
16N− 113
6+ 70
I If N = 150, then T =376
+ 70 = 7616◦F
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 37 / 54
. . . . . .
Solution
I The point-slope form of the equation for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equation
y− y0 = m(x− x0)
I The slope of our line is80− 70
173− 113=
1060
=16
I So an equation for T and N is
T− 70 =16(N− 113) =⇒ T =
16N− 113
6+ 70
I If N = 150, then T =376
+ 70 = 7616◦F
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 37 / 54
. . . . . .
Solution
I The point-slope form of the equation for a line is appropriatehere: If a line passes through (x0, y0) with slope m, then the linehas equation
y− y0 = m(x− x0)
I The slope of our line is80− 70
173− 113=
1060
=16
I So an equation for T and N is
T− 70 =16(N− 113) =⇒ T =
16N− 113
6+ 70
I If N = 150, then T =376
+ 70 = 7616◦F
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 37 / 54
. . . . . .
Other Polynomial functions
I Quadratic functions take the form
f(x) = ax2 + bx+ c
The graph is a parabola which opens upward if a > 0, downward ifa < 0.
I Cubic functions take the form
f(x) = ax3 + bx2 + cx+ d
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 38 / 54
. . . . . .
Other Polynomial functions
I Quadratic functions take the form
f(x) = ax2 + bx+ c
The graph is a parabola which opens upward if a > 0, downward ifa < 0.
I Cubic functions take the form
f(x) = ax3 + bx2 + cx+ d
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 38 / 54
. . . . . .
Example
A parabola passes through (0,3), (3,0), and (2,−1). What is theequation of the parabola?
SolutionThe general equation is y = ax2 + bx+ c. Each point gives anequation relating a, b, and c:
3 = a · 02 + b · 0+ c
−1 = a · 22 + b · 2+ c
0 = a · 32 + b · 3+ c
Right away we see c = 3. The other two equations become
−4 = 4a+ 2b−3 = 9a+ 3b
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 39 / 54
. . . . . .
Example
A parabola passes through (0,3), (3,0), and (2,−1). What is theequation of the parabola?
SolutionThe general equation is y = ax2 + bx+ c.
Each point gives anequation relating a, b, and c:
3 = a · 02 + b · 0+ c
−1 = a · 22 + b · 2+ c
0 = a · 32 + b · 3+ c
Right away we see c = 3. The other two equations become
−4 = 4a+ 2b−3 = 9a+ 3b
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 39 / 54
. . . . . .
Example
A parabola passes through (0,3), (3,0), and (2,−1). What is theequation of the parabola?
SolutionThe general equation is y = ax2 + bx+ c. Each point gives anequation relating a, b, and c:
3 = a · 02 + b · 0+ c
−1 = a · 22 + b · 2+ c
0 = a · 32 + b · 3+ c
Right away we see c = 3. The other two equations become
−4 = 4a+ 2b−3 = 9a+ 3b
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 39 / 54
. . . . . .
Example
A parabola passes through (0,3), (3,0), and (2,−1). What is theequation of the parabola?
SolutionThe general equation is y = ax2 + bx+ c. Each point gives anequation relating a, b, and c:
3 = a · 02 + b · 0+ c
−1 = a · 22 + b · 2+ c
0 = a · 32 + b · 3+ c
Right away we see c = 3. The other two equations become
−4 = 4a+ 2b−3 = 9a+ 3b
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 39 / 54
. . . . . .
Solution (Continued)
Multiplying the first equation by 3 and the second by 2 gives
−12 = 12a+ 6b−6 = 18a+ 6b
Subtract these two and we have −6 = −6a =⇒ a = 1. Substitutea = 1 into the first equation and we have
−12 = 12+ 6b =⇒ b = −4
So our equation isy = x2 − 4x+ 3
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 40 / 54
. . . . . .
Solution (Continued)
Multiplying the first equation by 3 and the second by 2 gives
−12 = 12a+ 6b−6 = 18a+ 6b
Subtract these two and we have −6 = −6a =⇒ a = 1.
Substitutea = 1 into the first equation and we have
−12 = 12+ 6b =⇒ b = −4
So our equation isy = x2 − 4x+ 3
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 40 / 54
. . . . . .
Solution (Continued)
Multiplying the first equation by 3 and the second by 2 gives
−12 = 12a+ 6b−6 = 18a+ 6b
Subtract these two and we have −6 = −6a =⇒ a = 1. Substitutea = 1 into the first equation and we have
−12 = 12+ 6b =⇒ b = −4
So our equation isy = x2 − 4x+ 3
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. . . . . .
Solution (Continued)
Multiplying the first equation by 3 and the second by 2 gives
−12 = 12a+ 6b−6 = 18a+ 6b
Subtract these two and we have −6 = −6a =⇒ a = 1. Substitutea = 1 into the first equation and we have
−12 = 12+ 6b =⇒ b = −4
So our equation isy = x2 − 4x+ 3
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 40 / 54
. . . . . .
Other power functions
I Whole number powers: f(x) = xn.
I negative powers are reciprocals: x−3 =1x3
.
I fractional powers are roots: x1/3 = 3√x.
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. . . . . .
Rational functions
DefinitionA rational function is a quotient of polynomials.
Example
The function f(x) =x3(x+ 3)
(x+ 2)(x− 1)is rational.
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. . . . . .
Trigonometric Functions
I Sine and cosineI Tangent and cotangentI Secant and cosecant
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. . . . . .
Exponential and Logarithmic functions
I exponential functions (for example f(x) = 2x)I logarithmic functions are their inverses (for example f(x) = log2(x))
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. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
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. . . . . .
Transformations of Functions
Take the squaring function and graph these transformations:
I y = (x+ 1)2
I y = (x− 1)2
I y = x2 + 1I y = x2 − 1
Observe that if the fiddling occurs within the function, a transformationis applied on the x-axis. After the function, to the y-axis.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 46 / 54
. . . . . .
Transformations of Functions
Take the squaring function and graph these transformations:
I y = (x+ 1)2
I y = (x− 1)2
I y = x2 + 1I y = x2 − 1
Observe that if the fiddling occurs within the function, a transformationis applied on the x-axis. After the function, to the y-axis.
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 46 / 54
. . . . . .
Vertical and Horizontal Shifts
Suppose c > 0. To obtain the graph ofI y = f(x) + c, shift the graph of y = f(x) a distance c units
upward
I y = f(x)− c, shift the graph of y = f(x) a distance c units
downward
I y = f(x−c), shift the graph of y = f(x) a distance c units
to the right
I y = f(x+ c), shift the graph of y = f(x) a distance c units
to the left
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. . . . . .
Vertical and Horizontal Shifts
Suppose c > 0. To obtain the graph ofI y = f(x) + c, shift the graph of y = f(x) a distance c units upwardI y = f(x)− c, shift the graph of y = f(x) a distance c units
downward
I y = f(x−c), shift the graph of y = f(x) a distance c units
to the right
I y = f(x+ c), shift the graph of y = f(x) a distance c units
to the left
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 47 / 54
. . . . . .
Vertical and Horizontal Shifts
Suppose c > 0. To obtain the graph ofI y = f(x) + c, shift the graph of y = f(x) a distance c units upwardI y = f(x)− c, shift the graph of y = f(x) a distance c units downwardI y = f(x−c), shift the graph of y = f(x) a distance c units
to the right
I y = f(x+ c), shift the graph of y = f(x) a distance c units
to the left
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 47 / 54
. . . . . .
Vertical and Horizontal Shifts
Suppose c > 0. To obtain the graph ofI y = f(x) + c, shift the graph of y = f(x) a distance c units upwardI y = f(x)− c, shift the graph of y = f(x) a distance c units downwardI y = f(x−c), shift the graph of y = f(x) a distance c units to the rightI y = f(x+ c), shift the graph of y = f(x) a distance c units
to the left
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 47 / 54
. . . . . .
Vertical and Horizontal Shifts
Suppose c > 0. To obtain the graph ofI y = f(x) + c, shift the graph of y = f(x) a distance c units upwardI y = f(x)− c, shift the graph of y = f(x) a distance c units downwardI y = f(x−c), shift the graph of y = f(x) a distance c units to the rightI y = f(x+ c), shift the graph of y = f(x) a distance c units to the left
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 47 / 54
. . . . . .
Now try these
I y = sin (2x)I y = 2 sin (x)I y = e−x
I y = −ex
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Scaling and flipping
To obtain the graph ofI y = f(c · x), scale the graph of f
horizontally
by cI y = c · f(x), scale the graph of f
vertically
by cI If |c| < 1, the scaling is a
compression
I If c < 0, the scaling includes a
flip
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. . . . . .
Scaling and flipping
To obtain the graph ofI y = f(c · x), scale the graph of f horizontally by cI y = c · f(x), scale the graph of f
vertically
by cI If |c| < 1, the scaling is a
compression
I If c < 0, the scaling includes a
flip
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 49 / 54
. . . . . .
Scaling and flipping
To obtain the graph ofI y = f(c · x), scale the graph of f horizontally by cI y = c · f(x), scale the graph of f vertically by cI If |c| < 1, the scaling is a
compression
I If c < 0, the scaling includes a
flip
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 49 / 54
. . . . . .
Scaling and flipping
To obtain the graph ofI y = f(c · x), scale the graph of f horizontally by cI y = c · f(x), scale the graph of f vertically by cI If |c| < 1, the scaling is a compressionI If c < 0, the scaling includes a
flip
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 49 / 54
. . . . . .
Scaling and flipping
To obtain the graph ofI y = f(c · x), scale the graph of f horizontally by cI y = c · f(x), scale the graph of f vertically by cI If |c| < 1, the scaling is a compressionI If c < 0, the scaling includes a flip
V63.0121.006/016, Calculus I (NYU) Section 1.1–1.2 Functions May 17, 2010 49 / 54
. . . . . .
Outline.
.
ModelingExamples of functions
Functions expressed by formulasFunctions described numericallyFunctions described graphicallyFunctions described verbally
Properties of functionsMonotonicitySymmetry
Classes of FunctionsLinear functionsOther Polynomial functionsOther power functionsRational functionsTrigonometric FunctionsExponential and Logarithmic functions
Transformations of FunctionsCompositions of Functions
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. . . . . .
Composition is a compounding of functions in
succession
..f .g
.g ◦ f
.x .(g ◦ f)(x).f(x)
.
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. . . . . .
Composing
Example
Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f.
Solutionf ◦ g(x) = sin2 x while g ◦ f(x) = sin(x2). Note they are not the same.
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. . . . . .
Composing
Example
Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f.
Solutionf ◦ g(x) = sin2 x while g ◦ f(x) = sin(x2). Note they are not the same.
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. . . . . .
Decomposing
Example
Express√x2 − 4 as a composition of two functions. What is its
domain?
SolutionWe can write the expression as f ◦ g, where f(u) =
√u and
g(x) = x2 − 4. The range of g needs to be within the domain of f. Toinsure that x2 − 4 ≥ 0, we must have x ≤ −2 or x ≥ 2.
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Summary
I The fundamental unit of investigation in calculus is the function.I Functions can have many representationsI There are many classes of algebraic functionsI Algebraic rules can be used to sketch graphs
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