Lecture 1 - Review of Functions.pdf
-
Upload
kent-harry-perez-cumpio -
Category
Documents
-
view
179 -
download
3
description
Transcript of Lecture 1 - Review of Functions.pdf
Review of Functions
Mathematics 53
Institute of Mathematics - UP Diliman
8 November 2012
Math 53 (Part 1) Review of Functions 8 November 2012 1 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 2 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 3 / 69
Functions
DefinitionLet X and Y be nonempty sets. A function f from X to Y, denoted f : X → Y, isa rule that assigns to each element of X a unique element of Y.
X: domain of f , denoted dom fY: codomain of fThe set of all elements of Y that are assigned to some element of X is therange of f , denoted ran f
Math 53 (Part 1) Review of Functions 8 November 2012 4 / 69
Functions
DefinitionLet X and Y be nonempty sets. A function f from X to Y, denoted f : X → Y, isa rule that assigns to each element of X a unique element of Y.
X: domain of f , denoted dom f
Y: codomain of fThe set of all elements of Y that are assigned to some element of X is therange of f , denoted ran f
Math 53 (Part 1) Review of Functions 8 November 2012 4 / 69
Functions
DefinitionLet X and Y be nonempty sets. A function f from X to Y, denoted f : X → Y, isa rule that assigns to each element of X a unique element of Y.
X: domain of f , denoted dom fY: codomain of f
The set of all elements of Y that are assigned to some element of X is therange of f , denoted ran f
Math 53 (Part 1) Review of Functions 8 November 2012 4 / 69
Functions
DefinitionLet X and Y be nonempty sets. A function f from X to Y, denoted f : X → Y, isa rule that assigns to each element of X a unique element of Y.
X: domain of f , denoted dom fY: codomain of fThe set of all elements of Y that are assigned to some element of X is therange of f , denoted ran f
Math 53 (Part 1) Review of Functions 8 November 2012 4 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f
= {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}
codomain: Y = {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y
= {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}
range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}range: ran f
= {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
−2 7−→ 4−1 7−→ 10 7−→ 01 7−→ 12 7−→ 4
domain: dom f = {−2,−1, 0, 1, 2}codomain: Y = {0, 1, 2, 3, 4}range: ran f = {0, 1, 4}
Math 53 (Part 1) Review of Functions 8 November 2012 5 / 69
Functions
If x ∈ X, the symbol f (x) denotes the element y ∈ Y that is assigned to x.
A function may be written as y = f (x)x: independent variable
y: dependent variable
Alternatively, a function f is a set of ordered pairs (x, y), where
(x, y) ∈ f if and only if y = f (x)
Math 53 (Part 1) Review of Functions 8 November 2012 6 / 69
Functions
If x ∈ X, the symbol f (x) denotes the element y ∈ Y that is assigned to x.
A function may be written as y = f (x)x: independent variable
y: dependent variable
Alternatively, a function f is a set of ordered pairs (x, y), where
(x, y) ∈ f if and only if y = f (x)
Math 53 (Part 1) Review of Functions 8 November 2012 6 / 69
Functions
If x ∈ X, the symbol f (x) denotes the element y ∈ Y that is assigned to x.
A function may be written as y = f (x)x: independent variable
y: dependent variable
Alternatively, a function f is a set of ordered pairs (x, y), where
(x, y) ∈ f if and only if y = f (x)
Math 53 (Part 1) Review of Functions 8 November 2012 6 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
The function f may be written as:
f (x) = x2 or y = x2
f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}
Math 53 (Part 1) Review of Functions 8 November 2012 7 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
The function f may be written as:
f (x) = x2 or y = x2
f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}
Math 53 (Part 1) Review of Functions 8 November 2012 7 / 69
Functions
ExampleConsider f : X → Y defined by the rule
x 7−→ x2
where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.
The function f may be written as:
f (x) = x2 or y = x2
f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}
Math 53 (Part 1) Review of Functions 8 November 2012 7 / 69
Real-valued functions of a single variable
Real-valued functions of a single variable:
Codomain: R
Math 53 deals with functions whose domain and range are subsets of R.
If the domain is not explicitly specified:
Domain: dom f = {x ∈ R | f (x) is a real number}
Math 53 (Part 1) Review of Functions 8 November 2012 8 / 69
Real-valued functions of a single variable
Real-valued functions of a single variable:
Codomain: R
Math 53 deals with functions whose domain and range are subsets of R.
If the domain is not explicitly specified:
Domain: dom f = {x ∈ R | f (x) is a real number}
Math 53 (Part 1) Review of Functions 8 November 2012 8 / 69
Real-valued functions of a single variable
Example
1 f (x) = x2
dom f = R
2 f (x) =x2 − 2x− 3
x + 1
dom f = R \ {−1}
Math 53 (Part 1) Review of Functions 8 November 2012 9 / 69
Real-valued functions of a single variable
Example
1 f (x) = x2
dom f = R
2 f (x) =x2 − 2x− 3
x + 1
dom f = R \ {−1}
Math 53 (Part 1) Review of Functions 8 November 2012 9 / 69
Real-valued functions of a single variable
Example
1 f (x) = x2
dom f = R
2 f (x) =x2 − 2x− 3
x + 1dom f = R \ {−1}
Math 53 (Part 1) Review of Functions 8 November 2012 9 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0
(x− 3)(x + 1) = 0x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Zeroes of a function
DefinitionA zero of a function f is a value of x for which f (x) = 0.
Example
Find the zero(es) of f (x) =x2 − 2x− 3
x + 1.
x2 − 2x− 3x + 1
= 0
x2 − 2x− 3 = 0(x− 3)(x + 1) = 0
x = 3 or x = −1
The only zero of f is x = 3.
Math 53 (Part 1) Review of Functions 8 November 2012 10 / 69
Graphs of Functions
DefinitionThe graph of a function f is the set of all points (x, y) in the plane R2 for which(x, y) ∈ f .
The graph of a function is the geometric representation on the Cartesian plane ofall points (x, y) that satisfy y = f (x).
Math 53 (Part 1) Review of Functions 8 November 2012 11 / 69
Graphs of Functions
DefinitionThe graph of a function f is the set of all points (x, y) in the plane R2 for which(x, y) ∈ f .
The graph of a function is the geometric representation on the Cartesian plane ofall points (x, y) that satisfy y = f (x).
Math 53 (Part 1) Review of Functions 8 November 2012 11 / 69
Graphs of Functions
Example
The graph of f (x) = x2:
−2 −1 1 2
1
2
3
4
0
The points on the graph of f are the points (x, y) that satisfy the equation y = x2.
Math 53 (Part 1) Review of Functions 8 November 2012 12 / 69
Graphs of Functions
Example
The graph of f (x) =x2 − 2x− 3
x + 1:
f (x) =x2 − 2x− 3
x + 1=
(x− 3)(x + 1)(x + 1)
= x− 3 if x 6= 1
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 13 / 69
Graphs of Functions
Example
The graph of f (x) =x2 − 2x− 3
x + 1:
f (x) =x2 − 2x− 3
x + 1
=(x− 3)(x + 1)
(x + 1)= x− 3 if x 6= 1
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 13 / 69
Graphs of Functions
Example
The graph of f (x) =x2 − 2x− 3
x + 1:
f (x) =x2 − 2x− 3
x + 1=
(x− 3)(x + 1)(x + 1)
= x− 3 if x 6= 1
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 13 / 69
Graphs of Functions
Example
The graph of f (x) =x2 − 2x− 3
x + 1:
f (x) =x2 − 2x− 3
x + 1=
(x− 3)(x + 1)(x + 1)
= x− 3 if x 6= 1
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 13 / 69
Graphs of Functions
Example
The graph of f (x) =x2 − 2x− 3
x + 1:
f (x) =x2 − 2x− 3
x + 1=
(x− 3)(x + 1)(x + 1)
= x− 3 if x 6= 1
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 13 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))
Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Graphs of Functions
Graphically:
Coordinates of a point on the graph in terms of x: (x, f (x))Domain: x-interval covered by the graph
Range: y-interval covered by the graph
Zero of a function: x-intercept of the graph
Intervals where the function value (or y-value) is positive: portions where thegraph lies above the x-axis
Intervals where the function value (or y-value) is negative: portions where thegraph lies below the x-axis
Math 53 (Part 1) Review of Functions 8 November 2012 14 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain:
R \ {−1}
Range:
R \ {−4}
Zero:
x = 3
Positive:
(3,+∞)
Negative:
(−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain: R \ {−1}Range:
R \ {−4}
Zero:
x = 3
Positive:
(3,+∞)
Negative:
(−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain: R \ {−1}Range: R \ {−4}Zero:
x = 3
Positive:
(3,+∞)
Negative:
(−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain: R \ {−1}Range: R \ {−4}Zero: x = 3
Positive:
(3,+∞)
Negative:
(−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain: R \ {−1}Range: R \ {−4}Zero: x = 3
Positive: (3,+∞)Negative:
(−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Consider the graph of f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
(x, x− 3)
Domain: R \ {−1}Range: R \ {−4}Zero: x = 3
Positive: (3,+∞)Negative: (−∞,−1) ∪ (−1, 3)
Math 53 (Part 1) Review of Functions 8 November 2012 15 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 16 / 69
Basic types of functions
Constant Functions - functions of the form f (x) = c, where c is a real number
dom f = R; ran f = {c}graph: horizontal line intersecting the y-axis at y = c
Math 53 (Part 1) Review of Functions 8 November 2012 17 / 69
Basic types of functions
Constant Functions - functions of the form f (x) = c, where c is a real number
dom f = R; ran f = {c}graph: horizontal line intersecting the y-axis at y = c
Math 53 (Part 1) Review of Functions 8 November 2012 17 / 69
Basic types of functions
ExampleConsider the constant function f (x) = 2.
−3 −2 −1 1 2 3
−1
1
2
3
0
Math 53 (Part 1) Review of Functions 8 November 2012 18 / 69
Basic types of functions
Linear Functions - functions of the form
f (x) = mx + b
with m 6= 0
dom f = R; ran f = R
graph: m is slope; y-intercept is b
Math 53 (Part 1) Review of Functions 8 November 2012 19 / 69
Basic types of functions
Linear Functions - functions of the form
f (x) = mx + b
with m 6= 0
dom f = R; ran f = R
graph: m is slope; y-intercept is b
Math 53 (Part 1) Review of Functions 8 November 2012 19 / 69
Basic types of functions
ExampleConsider the linear function f (x) = −x + 1.
−2 −1 1 2
−1
1
2
3
0
m = −1, y-intercept: 1
Math 53 (Part 1) Review of Functions 8 November 2012 20 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at
(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)
If a > 0: parabola opens upward, ran f =[
4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0:
parabola opens upward, ran f =[
4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward,
ran f =[
4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0:
parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Quadratic Functions - functions of the form
f (x) = ax2 + bx + c
with a 6= 0
dom f = R
graph: parabola with vertex at(− b
2a , 4ac−b2
4a
)If a > 0: parabola opens upward, ran f =
[4ac−b2
4a ,+∞)
If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2
4a
]
Math 53 (Part 1) Review of Functions 8 November 2012 21 / 69
Basic types of functions
Example
Consider the quadratic function f (x) = x2.
−2 −1 1 2
1
2
3
4
0
A parabola opening upward with vertex at (0, 0)
Math 53 (Part 1) Review of Functions 8 November 2012 22 / 69
Basic types of functions
Example
Consider the quadratic function f (x) = −x2 − 2x + 3.
−3 −2 −1 1 2
−2
−1
1
2
3
4
0
A parabola opening downward with vertex at (−1, 4)
Math 53 (Part 1) Review of Functions 8 November 2012 23 / 69
Basic types of functions
Extreme function values of a quadratic function:
a > 0: f has a minimum function valuea < 0: f has a maximum function value
The extreme function value of f occurs at x = − b2a and the extreme function
value of f is f(− b
2a
)= 4ac−b2
4a
Math 53 (Part 1) Review of Functions 8 November 2012 24 / 69
Basic types of functions
Example
Consider the quadratic function f (x) = −x2 − 2x + 3.
Since a < 0
f has a maximum function value which occurs at x = −1The maximum value of f (x) is f (−1) = 4
Math 53 (Part 1) Review of Functions 8 November 2012 25 / 69
Basic types of functions
Example
Consider the quadratic function f (x) = −x2 − 2x + 3.
Since a < 0
f has a maximum function value which occurs at x = −1The maximum value of f (x) is f (−1) = 4
Math 53 (Part 1) Review of Functions 8 November 2012 25 / 69
Basic types of functions
Example
Consider the quadratic function f (x) = −x2 − 2x + 3.
Since a < 0
f has a maximum function value which occurs at x = −1The maximum value of f (x) is f (−1) = 4
Math 53 (Part 1) Review of Functions 8 November 2012 25 / 69
Basic types of functions
Polynomial Functions - functions of the form
f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0
where n ∈W, an, an−1, ..., a0 are real numbers, with an 6= 0.
leading coefficient: an
degree of f : ndom f = R
Constant, linear and quadratic functions are special types of polynomialfunctions
Graphs of polynomial functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 26 / 69
Basic types of functions
Polynomial Functions - functions of the form
f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0
where n ∈W, an, an−1, ..., a0 are real numbers, with an 6= 0.
leading coefficient: an
degree of f : n
dom f = R
Constant, linear and quadratic functions are special types of polynomialfunctions
Graphs of polynomial functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 26 / 69
Basic types of functions
Polynomial Functions - functions of the form
f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0
where n ∈W, an, an−1, ..., a0 are real numbers, with an 6= 0.
leading coefficient: an
degree of f : ndom f = R
Constant, linear and quadratic functions are special types of polynomialfunctions
Graphs of polynomial functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 26 / 69
Basic types of functions
Polynomial Functions - functions of the form
f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0
where n ∈W, an, an−1, ..., a0 are real numbers, with an 6= 0.
leading coefficient: an
degree of f : ndom f = R
Constant, linear and quadratic functions are special types of polynomialfunctions
Graphs of polynomial functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 26 / 69
Basic types of functions
Polynomial Functions - functions of the form
f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0
where n ∈W, an, an−1, ..., a0 are real numbers, with an 6= 0.
leading coefficient: an
degree of f : ndom f = R
Constant, linear and quadratic functions are special types of polynomialfunctions
Graphs of polynomial functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 26 / 69
Basic types of functions
Rational Functions - functions of the form f (x) =p(x)q(x)
, where p and q are
polynomial functions, and q is not the constant zero function.
Domain: {x ∈ R | q(x) 6= 0}Graphs of rational functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 27 / 69
Basic types of functions
Rational Functions - functions of the form f (x) =p(x)q(x)
, where p and q are
polynomial functions, and q is not the constant zero function.
Domain: {x ∈ R | q(x) 6= 0}Graphs of rational functions: Unit 3
Math 53 (Part 1) Review of Functions 8 November 2012 27 / 69
Graphs of Functions
Example
Consider the rational function f (x) =x2 − 2x− 3
x + 1.
−3 −2 −1 1 2 3 4
−5
−4
−3
−2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 28 / 69
Basic types of functions
Functions involving rational exponents or radicals - functions of the form
f (x) = n√
x = x1/n
n is odd: dom f = R
n is even: dom f = [0, ∞)
Math 53 (Part 1) Review of Functions 8 November 2012 29 / 69
Basic types of functions
Functions involving rational exponents or radicals - functions of the form
f (x) = n√
x = x1/n
n is odd: dom f = R
n is even: dom f = [0, ∞)
Math 53 (Part 1) Review of Functions 8 November 2012 29 / 69
Basic types of functions
ExampleSquare root function: f (x) =
√x
y =√
x =⇒ y2 = x, y ≥ 0
1 2 3 4
−2
−1
1
2
0
The graph of x = y2
1 2 3 4
−2
−1
1
2
0
The graph of y =√
x
The graph of f (x) =√
x is the upper branch of the parabola x = y2
Math 53 (Part 1) Review of Functions 8 November 2012 30 / 69
Basic types of functions
ExampleSquare root function: f (x) =
√x
y =√
x =⇒ y2 = x, y ≥ 0
1 2 3 4
−2
−1
1
2
0
The graph of x = y2
1 2 3 4
−2
−1
1
2
0
The graph of y =√
x
The graph of f (x) =√
x is the upper branch of the parabola x = y2
Math 53 (Part 1) Review of Functions 8 November 2012 30 / 69
Basic types of functions
ExampleSquare root function: f (x) =
√x
y =√
x =⇒ y2 = x, y ≥ 0
1 2 3 4
−2
−1
1
2
0
The graph of x = y2
1 2 3 4
−2
−1
1
2
0
The graph of y =√
x
The graph of f (x) =√
x is the upper branch of the parabola x = y2
Math 53 (Part 1) Review of Functions 8 November 2012 30 / 69
Basic types of functions
Trigonometric/Circular Functions
sine, cosine, tangent, cotangent, secant and cosecant functions
In Math 53, the trigonometric functions are viewed as functions on the set ofreal numbers.
Math 53 (Part 1) Review of Functions 8 November 2012 31 / 69
Basic types of functions
Trigonometric/Circular Functions
sine, cosine, tangent, cotangent, secant and cosecant functions
In Math 53, the trigonometric functions are viewed as functions on the set ofreal numbers.
Math 53 (Part 1) Review of Functions 8 November 2012 31 / 69
Basic types of functions
Examplef (x) = sin x
−π−2π π 2π 3π 4π− π2− 3π
2π2
3π2
5π2
7π2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 32 / 69
Basic types of functions
Examplef (x) = cos x
−π−2π π 2π 3π 4π− π2− 3π
2π2
3π2
5π2
7π2
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 33 / 69
Basic Types of Functions
Examplef (x) = tan x
−π π 2π− π2− 3π
2π2
3π2
5π2
Math 53 (Part 1) Review of Functions 8 November 2012 34 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 35 / 69
Constructing a table of signs
The table of signs show when a given mathematical expression is positive, zero ornegative.
Two Methods:
1 Interval Method
2 Test Value Method
In both cases, one must determine the numbers where the given mathematicalexpression is zero or undefined.
Math 53 (Part 1) Review of Functions 8 November 2012 36 / 69
Constructing a table of signs
The table of signs show when a given mathematical expression is positive, zero ornegative.
Two Methods:
1 Interval Method
2 Test Value Method
In both cases, one must determine the numbers where the given mathematicalexpression is zero or undefined.
Math 53 (Part 1) Review of Functions 8 November 2012 36 / 69
Constructing a table of signs
The table of signs show when a given mathematical expression is positive, zero ornegative.
Two Methods:
1 Interval Method
2 Test Value Method
In both cases, one must determine the numbers where the given mathematicalexpression is zero or undefined.
Math 53 (Part 1) Review of Functions 8 November 2012 36 / 69
Constructing a table of signs
Example
Determine the intervals where the the graph of f (x) =2x2 − x3
2x2 − 3x + 1lies above
the x-axis.
We want to determine the intervals for which
2x2 − x3
2x2 − 3x + 1> 0
Math 53 (Part 1) Review of Functions 8 November 2012 37 / 69
Constructing a table of signs
Example
Determine the intervals where the the graph of f (x) =2x2 − x3
2x2 − 3x + 1lies above
the x-axis.
We want to determine the intervals for which
2x2 − x3
2x2 − 3x + 1
> 0
Math 53 (Part 1) Review of Functions 8 November 2012 37 / 69
Constructing a table of signs
Example
Determine the intervals where the the graph of f (x) =2x2 − x3
2x2 − 3x + 1lies above
the x-axis.
We want to determine the intervals for which
2x2 − x3
2x2 − 3x + 1> 0
Math 53 (Part 1) Review of Functions 8 November 2012 37 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1
=x2(2− x)
(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2
+ + + + +
2x− 1
− − + + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2
+ + + + +
2x− 1
− − + + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2
+ + + + +
2x− 1
− − + + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2
+ + + + +
2x− 1
− − + + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1
− − + + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − −
+ + +
x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1
− − − + +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − −
+ +
2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x
+ + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + +
−
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+
+ − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ +
− + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ + −
+ −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − +
−
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signsInterval Method: rewrite the expression as a product of factors whose table ofsigns are easily determined.
2x2 − x3
2x2 − 3x + 1=
x2(2− x)(2x− 1)(x− 1)
Zero at: x = 0, 2, Undefined at: x = 12 , 1
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −
x2(2− x)(2x− 1)(x− 1)
+ + − + −
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 38 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − + −
Sample point in (−∞, 0): x = −1
(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+
+ − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+
+ − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):
(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+
+ − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)
We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ +
− + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)
We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + −
+ −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)
We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − +
−
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)
We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)
We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Test Value Method: test a value in the specified interval
(−∞, 0)(
0, 12
) (12 , 1)
(1, 2) (2,+∞)
x2(2− x)(2x− 1)(x− 1)
+ + − + −
Sample point in (−∞, 0): x = −1(−1)2(3)(−3)(−2)
Sample point in(
0, 12
):(+)(+)
(−)(−)We get the same result:
The graph of f lies above the x-axis in the intervals (−∞, 0) ∪(
0, 12
)∪ (1, 2).
Math 53 (Part 1) Review of Functions 8 November 2012 39 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1
+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+
− + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ −
+ −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − +
−
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Constructing a table of signs
Example
Find the domain of f (x) =√−5x
x2 − 1.
Domain: x ∈ R such that−5x
x2 − 1≥ 0
Zero at: x = 0, Undefined at: x = −1, 1
(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x
x2 − 1+ − + −
Therefore,dom f = (∞,−1) ∪ [0, 1)
Math 53 (Part 1) Review of Functions 8 November 2012 40 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 41 / 69
Piecewise-defined functions
Piecewise-defined functions are functions that are defined by more than oneexpression. Such functions can be written in the form
f (x) =
f1(x) if x ∈ X1f2(x) if x ∈ X2
... if...
fn(x) if x ∈ Xn
where X1, ..., Xn ⊆ R with Xi ∩ Xj = ∅ for all i 6= j.
Math 53 (Part 1) Review of Functions 8 November 2012 42 / 69
Piecewise-defined functions
ExampleAn example of a piecewise function is the signum function (or sign function),denoted by sgn and defined by
sgn x =
−1 if x < 00 if x = 01 if x > 0
−3 −2 −1 1 2 3
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 43 / 69
Piecewise-defined functions
ExampleAn example of a piecewise function is the signum function (or sign function),denoted by sgn and defined by
sgn x =
−1 if x < 00 if x = 01 if x > 0
−3 −2 −1 1 2 3
−1
1
0
Math 53 (Part 1) Review of Functions 8 November 2012 43 / 69
The Absolute Value Function
Absolute Value Function - denoted by |x| and defined by
|x| =√
x2 =
{x, x ≥ 0−x, x < 0
Math 53 (Part 1) Review of Functions 8 November 2012 44 / 69
The Absolute Value Function
The graph of f (x) = |x|
−3 −2 −1 1 2 3
1
2
3
0
Math 53 (Part 1) Review of Functions 8 November 2012 45 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]]
= 2
2 [[2]]
= 2
3 [[0]]
= 0
4 [[−2.1]]
= −3
5 [[−π]]
= −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]] = 22 [[2]]
= 2
3 [[0]]
= 0
4 [[−2.1]]
= −3
5 [[−π]]
= −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]] = 22 [[2]] = 23 [[0]]
= 0
4 [[−2.1]]
= −3
5 [[−π]]
= −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]]
= −3
5 [[−π]]
= −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]] = −35 [[−π]]
= −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
Greatest Integer Function (GIF)
[[x]]: greatest integer less than or equal to x
Example
1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]] = −35 [[−π]] = −4
Math 53 (Part 1) Review of Functions 8 November 2012 46 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1,
− 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x
< 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 0
0, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00,
0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x
< 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 1
1, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 2
2, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3
......
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...
...
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...
...
In general,
[[x]] = n,
for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...
...
In general,
[[x]] = n, for
n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...
...
In general,
[[x]] = n, for n ≤ x
< n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer Function
As a piecewise function:
[[x]] =
......
− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...
...
In general,
[[x]] = n, for n ≤ x < n + 1 where n ∈ Z
Math 53 (Part 1) Review of Functions 8 November 2012 47 / 69
The Greatest Integer FunctionThe graph of f (x) = [[x]]
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 48 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 49 / 69
Operations on Functions
Definition (Operations on Functions)Let f and g be functions, c ∈ R.
1 Addition: ( f + g)(x) = f (x) + g(x); dom( f + g) = dom f ∩ dom g2 Subtraction: ( f − g)(x) = f (x)− g(x); dom( f − g) = dom f ∩ dom g3 Multiplication: ( f g)(x) = f (x)g(x); dom( f g) = dom f ∩ dom g
4 Division:(
fg
)(x) =
f (x)g(x)
;
dom(
fg
)= (dom f ∩ dom g) \ {x ∈ dom g | g(x) = 0}
5 Composition: ( f ◦ g)(x) = f (g(x));dom( f ◦ g) = {x ∈ dom g | g(x) ∈ dom f }
6 Scalar Multiplication: c f (x) = c ( f (x)); dom c f = dom f
Math 53 (Part 1) Review of Functions 8 November 2012 50 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin xh(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin xh(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin xh(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin x
h(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin xh(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.
Let
f (x) = x2
g(x) = sin xh(x) = 3x− 1
ThenF(x) = ( f ◦ g ◦ h) (x)
Math 53 (Part 1) Review of Functions 8 November 2012 51 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)]
=1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h
= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = x2 and g(x) = x + h. Find1h[( f ◦ g) (x)− f (x)].
1h[( f ◦ g) (x)− f (x)] =
1h[ f (g(x))− f (x)]
=f (x + h)− f (x)
h
=(x + h)2 − x2
h
=(x2 + 2xh + h2)− x2
h
=2xh + h2
h= 2x + h
Math 53 (Part 1) Review of Functions 8 November 2012 52 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
Recall:
|x| =√
x2 =
{x, x ≥ 0−x, x < 0
Therefore,
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
Math 53 (Part 1) Review of Functions 8 November 2012 53 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
Recall:
|x| =√
x2 =
{x, x ≥ 0−x, x < 0
Therefore,
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
Math 53 (Part 1) Review of Functions 8 November 2012 53 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
Recall:
|x| =√
x2 =
{x, x ≥ 0−x, x < 0
Therefore,
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
Math 53 (Part 1) Review of Functions 8 November 2012 53 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
(−∞,−1) (−1, 1) (1,+∞)
x− 1 − − +
x + 1 − + +
x2 − 1 + − +
( f ◦ g) (x) = |x2 − 1| ={
x2 − 1, x ≥ 1 or x ≤ 11− x2, −1 < x < 1
Math 53 (Part 1) Review of Functions 8 November 2012 54 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
(−∞,−1) (−1, 1) (1,+∞)
x− 1 − − +
x + 1 − + +
x2 − 1 + − +
( f ◦ g) (x) = |x2 − 1| ={
x2 − 1, x ≥ 1 or x ≤ 11− x2, −1 < x < 1
Math 53 (Part 1) Review of Functions 8 November 2012 54 / 69
Operations on Functions
Example
Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.
( f ◦ g) (x) = f (g(x)) = |x2 − 1| ={
x2 − 1, x2 − 1 ≥ 0−(x2 − 1), x2 − 1 < 0
(−∞,−1) (−1, 1) (1,+∞)
x− 1 − − +
x + 1 − + +
x2 − 1 + − +
( f ◦ g) (x) = |x2 − 1| ={
x2 − 1, x ≥ 1 or x ≤ 11− x2, −1 < x < 1
Math 53 (Part 1) Review of Functions 8 November 2012 54 / 69
( f ◦ g) (x) = |x2 − 1| ={
x2 − 1, x ≥ 1 or x ≤ 11− x2, −1 < x < 1
−3 −2 −1 1 2 3
1
2
3
Math 53 (Part 1) Review of Functions 8 November 2012 55 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1
2≤ x <
n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n
if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1
2≤ x <
n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1
2≤ x <
n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1
n− 1 ≤ 2x < nn− 1
2≤ x <
n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < n
n− 12
≤ x <n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1
n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1
2≤ x <
n2
Math 53 (Part 1) Review of Functions 8 November 2012 56 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1
2≤ x <
n2
[[2x + 1]] =
......
−1, if −1 ≤ x < − 12
0, if − 12 ≤ x < 0
1, if 0 ≤ x < 12
2, if 12 ≤ x < 1
......
Math 53 (Part 1) Review of Functions 8 November 2012 57 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1
2≤ x <
n2
[[2x + 1]] =
......
−1, if −1 ≤ x < − 12
0, if − 12 ≤ x < 0
1, if 0 ≤ x < 12
2, if 12 ≤ x < 1
......
Math 53 (Part 1) Review of Functions 8 November 2012 57 / 69
Operations on Functions
ExampleLet f (x) = 2x + 1 and g(x) = [[x]]. Express (g ◦ f ) (x) as a piecewise function.
(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1
2≤ x <
n2
[[2x + 1]] =
......
−1, if −1 ≤ x < − 12
0, if − 12 ≤ x < 0
1, if 0 ≤ x < 12
2, if 12 ≤ x < 1
......
Math 53 (Part 1) Review of Functions 8 November 2012 57 / 69
−4 −3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 58 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
y =√
2− x
y2 = 2− x, y ≥ 0
x = 2− y2, y ≥ 0
The graph of g is the upper branch of the parabola x = 2− y2.
Math 53 (Part 1) Review of Functions 8 November 2012 59 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
y =√
2− x
y2 = 2− x, y ≥ 0
x = 2− y2, y ≥ 0
The graph of g is the upper branch of the parabola x = 2− y2.
Math 53 (Part 1) Review of Functions 8 November 2012 59 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
y =√
2− x
y2 = 2− x, y ≥ 0
x = 2− y2, y ≥ 0
The graph of g is the upper branch of the parabola x = 2− y2.
Math 53 (Part 1) Review of Functions 8 November 2012 59 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
y =√
2− x
y2 = 2− x, y ≥ 0
x = 2− y2, y ≥ 0
The graph of g is the upper branch of the parabola x = 2− y2.
Math 53 (Part 1) Review of Functions 8 November 2012 59 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
y =√
2− x
y2 = 2− x, y ≥ 0
x = 2− y2, y ≥ 0
The graph of g is the upper branch of the parabola x = 2− y2.
Math 53 (Part 1) Review of Functions 8 November 2012 59 / 69
Other Graphing Examples
Example
Graph g(x) =√
2− x.
−2 −1 1 2
−2
−1
1
2
0
Math 53 (Part 1) Review of Functions 8 November 2012 60 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
y = −√
4− x2
y2 = 4− x2, y ≤ 0
x2 + y2 = 4, y ≤ 0
The graph of h is the lower semicircle of the circle x2 + y2 = 4.
Math 53 (Part 1) Review of Functions 8 November 2012 61 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
y = −√
4− x2
y2 = 4− x2, y ≤ 0
x2 + y2 = 4, y ≤ 0
The graph of h is the lower semicircle of the circle x2 + y2 = 4.
Math 53 (Part 1) Review of Functions 8 November 2012 61 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
y = −√
4− x2
y2 = 4− x2, y ≤ 0
x2 + y2 = 4, y ≤ 0
The graph of h is the lower semicircle of the circle x2 + y2 = 4.
Math 53 (Part 1) Review of Functions 8 November 2012 61 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
y = −√
4− x2
y2 = 4− x2, y ≤ 0
x2 + y2 = 4, y ≤ 0
The graph of h is the lower semicircle of the circle x2 + y2 = 4.
Math 53 (Part 1) Review of Functions 8 November 2012 61 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
y = −√
4− x2
y2 = 4− x2, y ≤ 0
x2 + y2 = 4, y ≤ 0
The graph of h is the lower semicircle of the circle x2 + y2 = 4.
Math 53 (Part 1) Review of Functions 8 November 2012 61 / 69
Other Graphing Examples
Example
Graph h(x) = −√
4− x2.
−2 −1 1 2
−2
−1
1
2
Math 53 (Part 1) Review of Functions 8 November 2012 62 / 69
Other Graphing Examples
Example
Graph f (x) =
x + 4 if x < −2
x3 + x2
x + 1if −2 ≤ x ≤ 2
|x− 6| if x > 2
.
f (x) =
x + 4 if x < −2
x2(x + 1)x + 1
if −2 ≤ x ≤ 2
−(x− 6) if 2 < x < 6x− 6 if x ≥ 6
f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
Math 53 (Part 1) Review of Functions 8 November 2012 63 / 69
Other Graphing Examples
Example
Graph f (x) =
x + 4 if x < −2
x3 + x2
x + 1if −2 ≤ x ≤ 2
|x− 6| if x > 2
.
f (x) =
x + 4 if x < −2
x2(x + 1)x + 1
if −2 ≤ x ≤ 2
−(x− 6) if 2 < x < 6x− 6 if x ≥ 6
f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
Math 53 (Part 1) Review of Functions 8 November 2012 63 / 69
Other Graphing Examples
Example
Graph f (x) =
x + 4 if x < −2
x3 + x2
x + 1if −2 ≤ x ≤ 2
|x− 6| if x > 2
.
f (x) =
x + 4 if x < −2
x2(x + 1)x + 1
if −2 ≤ x ≤ 2
−(x− 6) if 2 < x < 6x− 6 if x ≥ 6
f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
Math 53 (Part 1) Review of Functions 8 November 2012 63 / 69
Other Graphing Examples
The graph of f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 64 / 69
Other Graphing Examples
The graph of f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 64 / 69
Other Graphing Examples
The graph of f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 64 / 69
Other Graphing Examples
The graph of f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 64 / 69
Other Graphing Examples
The graph of f (x) =
x + 4 if x < −2
x2 if −2 ≤ x ≤ 2, x 6= −1−x + 6 if 2 < x < 6x− 6 if x ≥ 6
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10−1
1
2
3
4
0
Math 53 (Part 1) Review of Functions 8 November 2012 64 / 69
Outline
1 Functions
2 Basic Types of Functions
3 Constructing a table of signs
4 Piecewise-defined functions
5 Operations on Functions
6 Functions as Mathematical Models
Math 53 (Part 1) Review of Functions 8 November 2012 65 / 69
Functions as Mathematical Models
Express a certain situation as a functional relationship between certainquantities
Math 53 (Part 1) Review of Functions 8 November 2012 66 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the field
The area A of the field:A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240
y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleA rectangular field has a perimeter of 240 meters. Express the area of the field asa function of its width.
Let x be the width and y be the length of the fieldThe area A of the field:
A = xy
Since the perimeter of the field is 240 meters:
2x + 2y = 240y = 120− x
The area of the field expressed as a function of x:
A(x) = x(120− x) = −x2 + 120x
Math 53 (Part 1) Review of Functions 8 November 2012 67 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.
Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.
Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.
Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.
Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.
Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.
Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Functions as Mathematical Models
ExampleFind two numbers whose difference is 14 and whose product is minimum.
Let x be the greater number and y be the smaller number.Since the difference of 14 is positive
x− y = 14y = x− 14
The product as a function of x is
P(x) = x(x− 14) = x2 − 14x
P is a quadratic function with a minimum function value at
x = − b2a
= 7
The two numbers are 7 and −7, and the minimum product is P(7) = −49.Math 53 (Part 1) Review of Functions 8 November 2012 68 / 69
Announcements
Unit 1 Module will be available on 16 November 2012.
Google site: https://sites.google.com/a/math.upd.edu.ph/m53-s2-1213
All lecture slides will be posted in the website.
A printer–friendly, condensed version of slides for the 1st three lectures willbe uploaded in the website prior to the lecture.
Math 53 (Part 1) Review of Functions 8 November 2012 69 / 69