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

Outline

1 Functions

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

Functions

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

Functions

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

Functions

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

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}

Functions

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

Functions

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}

Functions

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}

Functions

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}

Functions

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}

Functions

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}

Functions

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

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)

Functions

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)}

Functions

x 7−→ x2

where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.

f (x) = x2 or y = x2

f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}

Functions

x 7−→ x2

where X = {−2,−1, 0, 1, 2} and Y = {0, 1, 2, 3, 4}.

f (x) = x2 or y = x2

f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4)}

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}

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}

Example

1 f (x) = x2

dom f = R

2 f (x) =x2 − 2x− 3

dom f = R \ {−1}

Example

1 f (x) = x2

dom f = R

2 f (x) =x2 − 2x− 3

dom f = R \ {−1}

Example

1 f (x) = x2

dom f = R

2 f (x) =x2 − 2x− 3

x + 1dom f = R \ {−1}

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

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

The only zero of f is x = 3.

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0

(x− 3)(x + 1) = 0x = 3 or x = −1

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

Example

x + 1.

x2 − 2x− 3x + 1

x2 − 2x− 3 = 0(x− 3)(x + 1) = 0

x = 3 or x = −1

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).

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).

Graphs of Functions

Example

The graph of f (x) = x2:

−2 −1 1 2

The points on the graph of f are the points (x, y) that satisfy the equation y = x2.

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

Graphs of Functions

Example

x + 1:

f (x) =x2 − 2x− 3

=(x− 3)(x + 1)

(x + 1)= x− 3 if x 6= 1

−3 −2 −1 1 2 3 4

Graphs of Functions

Example

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

Graphs of Functions

Example

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

Graphs of Functions

Example

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

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

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

Graphs of Functions

Graphically:

Graphs of Functions

Graphically:

Graphs of Functions

Graphically:

Graphs of Functions

Graphically:

Graphs of Functions

Graphically:

Consider the graph of f (x) =x2 − 2x− 3

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Domain:

R \ {−1}

Range:

R \ {−4}

Positive:

(3,+∞)

Negative:

(−∞,−1) ∪ (−1, 3)

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Domain: R \ {−1}Range:

R \ {−4}

Positive:

(3,+∞)

Negative:

(−∞,−1) ∪ (−1, 3)

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Domain: R \ {−1}Range: R \ {−4}Zero:

Positive:

(3,+∞)

Negative:

(−∞,−1) ∪ (−1, 3)

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Domain: R \ {−1}Range: R \ {−4}Zero: x = 3

Positive:

(3,+∞)

Negative:

(−∞,−1) ∪ (−1, 3)

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Positive: (3,+∞)Negative:

(−∞,−1) ∪ (−1, 3)

x + 1.

−3 −2 −1 1 2 3 4

(x, x− 3)

Positive: (3,+∞)Negative: (−∞,−1) ∪ (−1, 3)

Outline

1 Functions

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

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

ExampleConsider the constant function f (x) = 2.

−3 −2 −1 1 2 3

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

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

ExampleConsider the linear function f (x) = −x + 1.

−2 −1 1 2

m = −1, y-intercept: 1

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

)If a > 0: parabola opens upward, ran f =

[4ac−b2

4a ,+∞)

If a < 0: parabola opens downward, ran f =(−∞, 4ac−b2

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

[4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

graph: parabola with vertex at

(− b

2a , 4ac−b2

[4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

If a > 0: parabola opens upward, ran f =[

4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

)If a > 0:

parabola opens upward, ran f =[

4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

)If a > 0: parabola opens upward,

ran f =[

4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

[4ac−b2

4a ,+∞)

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

[4ac−b2

4a ,+∞)

If a < 0:

parabola opens downward, ran f =(−∞, 4ac−b2

f (x) = ax2 + bx + c

with a 6= 0

dom f = R

2a , 4ac−b2

[4ac−b2

4a ,+∞)

Example

Consider the quadratic function f (x) = x2.

−2 −1 1 2

A parabola opening upward with vertex at (0, 0)

Example

Consider the quadratic function f (x) = −x2 − 2x + 3.

−3 −2 −1 1 2

A parabola opening downward with vertex at (−1, 4)

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

)= 4ac−b2

Example

Since a < 0

f has a maximum function value which occurs at x = −1The maximum value of f (x) is f (−1) = 4

Example

Since a < 0

Example

Since a < 0

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

f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0

degree of f : n

dom f = R

f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0

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

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

Graphs of Functions

Example

Consider the rational function f (x) =x2 − 2x− 3

x + 1.

−3 −2 −1 1 2 3 4

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, ∞)

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, ∞)

ExampleSquare root function: f (x) =

y =√

x =⇒ y2 = x, y ≥ 0

1 2 3 4

The graph of x = y2

1 2 3 4

The graph of y =√

The graph of f (x) =√

x is the upper branch of the parabola x = y2

y =√

x =⇒ y2 = x, y ≥ 0

1 2 3 4

The graph of x = y2

1 2 3 4

The graph of y =√

y =√

x =⇒ y2 = x, y ≥ 0

1 2 3 4

The graph of x = y2

1 2 3 4

The graph of y =√

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.

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.

Examplef (x) = sin x

−π−2π π 2π 3π 4π− π2− 3π

Examplef (x) = cos x

−π−2π π 2π 3π 4π− π2− 3π

Basic Types of Functions

Examplef (x) = tan x

−π π 2π− π2− 3π

Outline

1 Functions

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.

Two Methods:

1 Interval Method

2 Test Value Method

Two Methods:

1 Interval Method

2 Test Value Method

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

Example

the x-axis.

2x2 − x3

2x2 − 3x + 1

Example

the x-axis.

2x2 − x3

2x2 − 3x + 1> 0

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)(

) (12 , 1)

(1, 2) (2,+∞)

+ + + + +

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) ∪(

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

+ + + + +

2x− 1

− − + + +

x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

+ + + + +

2x− 1

− − + + +

x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

+ + + + +

2x− 1

− − + + +

x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1

− − + + +

x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − −

x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1

− − − + +

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − −

2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x

+ + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + +

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

− + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − +

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

2x2 − x3

2x2 − 3x + 1=

x2(2− x)(2x− 1)(x− 1)

(−∞, 0)(

) (12 , 1)

(1, 2) (2,+∞)

x2 + + + + +2x− 1 − − + + +x− 1 − − − + +2− x + + + + −

x2(2− x)(2x− 1)(x− 1)

+ + − + −

)∪ (1, 2).

Test Value Method: test a value in the specified interval

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)We get the same result:

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

)∪ (1, 2).

(−∞, 0)(

) (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(

(+)(+)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)

We get the same result:

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

(−)(−)

)∪ (1, 2).

(−∞, 0)(

) (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(

):(+)(+)

)∪ (1, 2).

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)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ − + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ − + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1

+ − + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+

− + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ − +

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ − + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Example

x2 − 1.

x2 − 1≥ 0

(−∞,−1) (−1, 0) (0, 1) (1,+∞)−5x

x2 − 1+ − + −

Therefore,dom f = (∞,−1) ∪ [0, 1)

Outline

1 Functions

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.

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

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

The Absolute Value Function

Absolute Value Function - denoted by |x| and defined by

|x| =√

{x, x ≥ 0−x, x < 0

The Absolute Value Function

The graph of f (x) = |x|

−3 −2 −1 1 2 3

The Greatest Integer Function

Greatest Integer Function (GIF)

[[x]]: greatest integer less than or equal to x

Example

1 [[2.4]]

2 [[2]]

3 [[0]]

4 [[−2.1]]

= −3

5 [[−π]]

= −4

Example

1 [[2.4]] = 22 [[2]]

3 [[0]]

4 [[−2.1]]

= −3

5 [[−π]]

= −4

Example

1 [[2.4]] = 22 [[2]] = 23 [[0]]

4 [[−2.1]]

= −3

5 [[−π]]

= −4

Example

1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]]

= −3

5 [[−π]]

= −4

Example

1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]] = −35 [[−π]]

= −4

Example

1 [[2.4]] = 22 [[2]] = 23 [[0]] = 04 [[−2.1]] = −35 [[−π]] = −4

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

[[x]] =

......

− 1,

− 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x

< 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 0

0, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00,

0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x

< 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x < 1

1, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 2

2, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3

......

In general,

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...

In general,

[[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

[[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

[[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

[[x]] =

......

− 1, − 1 ≤ x < 00, 0 ≤ x < 11, 1 ≤ x < 22, 2 ≤ x < 3...

In general,

The Greatest Integer FunctionThe graph of f (x) = [[x]]

−4 −3 −2 −1 1 2 3 4

Outline

1 Functions

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:(

)(x) =

f (x)g(x)

)= (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

Example

Express the function F(x) = sin2 (3x− 1) as a composition of three functionslisted among the basic types of functions.

f (x) = x2

g(x) = sin xh(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

Example

f (x) = x2

g(x) = sin xh(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

Example

f (x) = x2

g(x) = sin xh(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

Example

f (x) = x2

g(x) = sin x

h(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

Example

f (x) = x2

g(x) = sin xh(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

Example

f (x) = x2

g(x) = sin xh(x) = 3x− 1

ThenF(x) = ( f ◦ g ◦ h) (x)

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)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)]

=1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

= 2x + h

Example

1h[( f ◦ g) (x)− f (x)] =

1h[ f (g(x))− f (x)]

=f (x + h)− f (x)

=(x + h)2 − x2

=(x2 + 2xh + h2)− x2

=2xh + h2

h= 2x + h

Example

Let f (x) = |x| and g(x) = x2 − 1. Express ( f ◦ g) (x) as a piecewise function.

Recall:

|x| =√

{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

Example

Recall:

|x| =√

{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

Example

Recall:

|x| =√

{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

Example

( 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

Example

( 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

Example

( 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

( f ◦ g) (x) = |x2 − 1| ={

x2 − 1, x ≥ 1 or x ≤ 11− x2, −1 < x < 1

−3 −2 −1 1 2 3

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 <

(g ◦ f ) (x) = [[2x + 1]] = n

if n ≤ 2x + 1 < n + 1

n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1

2≤ x <

(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1

n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1

2≤ x <

(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1

n ≤ 2x + 1 < n + 1

n− 1 ≤ 2x < nn− 1

2≤ x <

(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1

n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < n

n− 12

≤ x <n2

(g ◦ f ) (x) = [[2x + 1]] = n if n ≤ 2x + 1 < n + 1

n ≤ 2x + 1 < n + 1n− 1 ≤ 2x < nn− 1

2≤ x <

(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1

2≤ x <

[[2x + 1]] =

......

−1, if −1 ≤ x < − 12

0, if − 12 ≤ x < 0

1, if 0 ≤ x < 12

2, if 12 ≤ x < 1

......

(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1

2≤ x <

[[2x + 1]] =

......

−1, if −1 ≤ x < − 12

0, if − 12 ≤ x < 0

1, if 0 ≤ x < 12

2, if 12 ≤ x < 1

......

(g ◦ f ) (x) = [[2x + 1]] = n, ifn− 1

2≤ x <

[[2x + 1]] =

......

−1, if −1 ≤ x < − 12

0, if − 12 ≤ x < 0

1, if 0 ≤ x < 12

2, if 12 ≤ x < 1

......

−4 −3 −2 −1 1 2 3 4

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.

Example

Graph g(x) =√

2− x.

y =√

2− x

y2 = 2− x, y ≥ 0

x = 2− y2, y ≥ 0

Example

Graph g(x) =√

2− x.

y =√

2− x

y2 = 2− x, y ≥ 0

x = 2− y2, y ≥ 0

Example

Graph g(x) =√

2− x.

y =√

2− x

y2 = 2− x, y ≥ 0

x = 2− y2, y ≥ 0

Example

Graph g(x) =√

2− x.

y =√

2− x

y2 = 2− x, y ≥ 0

x = 2− y2, y ≥ 0

Example

Graph g(x) =√

2− x.

−2 −1 1 2

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.

Example

Graph h(x) = −√

4− x2.

y = −√

4− x2

y2 = 4− x2, y ≤ 0

x2 + y2 = 4, y ≤ 0

Example

Graph h(x) = −√

4− x2.

y = −√

4− x2

y2 = 4− x2, y ≤ 0

x2 + y2 = 4, y ≤ 0

Example

Graph h(x) = −√

4− x2.

y = −√

4− x2

y2 = 4− x2, y ≤ 0

x2 + y2 = 4, y ≤ 0

Example

Graph h(x) = −√

4− x2.

y = −√

4− x2

y2 = 4− x2, y ≤ 0

x2 + y2 = 4, y ≤ 0

Example

Graph h(x) = −√

4− x2.

−2 −1 1 2

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

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

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

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

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

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

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

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

Outline

1 Functions

Functions as Mathematical Models

Express a certain situation as a functional relationship between certainquantities

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

Let x be the width and y be the length of the field

The area A of the field:A = xy

2x + 2y = 240y = 120− x

A(x) = x(120− x) = −x2 + 120x

A = xy

2x + 2y = 240y = 120− x

A(x) = x(120− x) = −x2 + 120x

A = xy

2x + 2y = 240y = 120− x

A(x) = x(120− x) = −x2 + 120x

A = xy

2x + 2y = 240

y = 120− x

A(x) = x(120− x) = −x2 + 120x

A = xy

2x + 2y = 240y = 120− x

A(x) = x(120− x) = −x2 + 120x

A = xy

2x + 2y = 240y = 120− x

A(x) = x(120− x) = −x2 + 120x

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

The two numbers are 7 and −7, and the minimum product is P(7) = −49.

Let x be the greater number and y be the smaller number.

Since the difference of 14 is positive

x− y = 14y = x− 14

P(x) = x(x− 14) = x2 − 14x

x = − b2a

x− y = 14y = x− 14

P(x) = x(x− 14) = x2 − 14x

x = − b2a

x− y = 14y = x− 14

P(x) = x(x− 14) = x2 − 14x

x = − b2a

x− y = 14y = x− 14

P(x) = x(x− 14) = x2 − 14x

x = − b2a

x− y = 14y = x− 14

P(x) = x(x− 14) = x2 − 14x

x = − b2a

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.

lecture 1 - review of functions.pdf

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