M14 Relations and Functions.pdf

Functions: Domain,

Range, Operations

At the end of this lecture, a student must be able to:

1 _{Illustrate the difference of relations from functions}

2 _{Determine the domain, codomain, and range of a function}

3 _{Interpret the notationf(x)}

4 _{Find the (natural) domain of a function f when given}

expressionf(x)

Relations

Recall: X×Y ={(x, y)|x∈X and y ∈Y}

Definition

A relationis a non-empty set of ordered pairs.

Relations

Recall: X×Y ={(x, y)|x∈X and y ∈Y}

Definition

A relationis a non-empty set of ordered pairs.

Example: Let X ={x1, x2, x3, x4}, Y ={y1, y2, y3, y4},

R ={(x2, y1),(x2, y2),(x3, y3)}

X

x1

x2

x4

x3

R

Y y1

y2

y4

y3

R is a relation from X toY since R ⊆X×Y, and R6=_∅.

Example: Let X ={x1, x2, x3, x4}, Y ={y1, y2, y3, y4},

R ={(x2, y1),(x2, y2),(x3, y3)}

X

x1

x2

x4

x3

R

Y y1

y2

y4

y3

R is a relation from X toY since R ⊆X×Y, and R6=_∅.

Applications of Functions

- describing a quantity in terms of another,

Bus fare depends on distance traveled

Cost of production is a function of the quantity on raw material used

Applications of Functions

- describing a quantity in terms of another, Bus fare depends on distance traveled

Cost of production is a function of the quantity on raw material used

Functions

Definition

A function from X toY, denoted f :X −→Y, is a relation from X to Y such that for every x∈X, there is a unique y ∈Y such that (x, y)∈f.

Example: Do we have a function?

1. X ={x1, x2, x3, x4}, Y ={y1, y2, y3, y4}

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f =

Example: Do we have a function?

1. X ={x1, x2, x3, x4}, Y ={y1, y2, y3, y4}

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

2. A={1,2,3,4,5}, B =_N g={(x, y)∈A×B |y= 2x}

={(1,2),(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),

(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),(2,4),

(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

2. A={1,2,3,4,5}, B =_N

g={(x, y)∈A×B |y= 2x} ={(1,2),(2,4),(3,6),(4,8),(5,10)}

3.

X

x1

x2

x4

x3

R Y

y1

y2

y4

y3

R =

{(x2, y1),(x2, y2),(x3, y3)}

R is not a function

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)}

Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}

No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1}

No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1} No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each x in}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1} No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2}}

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1} No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each xin}

R, there corresponds only one y,

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1} No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each xin}

R, there corresponds only one y, y=x2+ 2,

Determine whether or not the following represent functions.

1. f ={(0,1),(2,3),(4,3)} Yes.

2. g ={(0,1),(2,3),(2,4)}No.

3. h={(x, y)∈_R×_R|x2+y2 = 1} No, because (0,1)∈h and (0,−1)∈h.

4. p={(x, y)∈_R×_R|y−x2 _{= 2} Yes, because for each xin}

We can also think of a function as a machine which processes an input and gives exactly one output for each input.

In a way, the value of an output y depends on an input value x. Thus,x is called theindependent variable and y is called the

We can also think of a function as a machine which processes an input and gives exactly one output for each input.

In a way, the value of an output y depends on an input value x.

Thus,x is called theindependent variable and y is called the

We can also think of a function as a machine which processes an input and gives exactly one output for each input.

In a way, the value of an output y depends on an input value x. Thus,x is called theindependent variable and y is called the

Whenever (x, y)∈f, we write

y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =

y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =

y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Whenever (x, y)∈f, we write y=f(x).

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

X ={x1, x2, x3, x4},

Y ={y1, y2, y3, y4}

f(x1) =y1,

f(x2) =y2,

f(x3) =y2,

f(x4) = y1

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 range of f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf.

2 Y: codomain of f, denoted codomf.

3 range of f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 range of f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated.

That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y domf =

{x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf =

{y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

ranf =

Letf be a function from X toY.

1 X: domain of f, denoted domf. 2 Y: codomain of f, denoted codomf.

3 rangeof f, denoted ranf: set of all y∈Y to which some

x∈X is associated. That is,

ranf ={y∈Y |y=f(x) for some x∈X}.

X

x1

x2

x4

x3

f Y

y1

y2

y4

y3

f :X −→Y

domf ={x1, x2, x3, x4},

codomf ={y1, y2, y3, y4},

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x

g(1) = 2, g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1)

= 2, g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4,

g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

Example: Let A={1,2,3,4,5}, B =_Nand g :A −→B, such that for eachx∈A, g(x) = 2x g(1) = 2,

g(2) = 4, g(3) = 6, g(4) = 8, g(5) = 10

domg =A, codomg =B,

rang ={2,4,6,8,10}

When given f(x) = an expression involving the variable x, this gives us a rule to obtain the images of elements of the domain.

Example:

If the functionf squares a real number, thenf(x) =x2_.

When given f(x) = an expression involving the variable x, this gives us a rule to obtain the images of elements of the domain.

Example:

If the functionf squares a real number,

then f(x) =x2_.

When given f(x) = an expression involving the variable x, this gives us a rule to obtain the images of elements of the domain.

Example:

If the functionf squares a real number, thenf(x) =x2_.

When given f(x) = an expression involving the variable x, this gives us a rule to obtain the images of elements of the domain.

Example:

If the functionf squares a real number, thenf(x) =x2_.

Ifg(x) adds 1 to a positive number,

When given f(x) = an expression involving the variable x, this gives us a rule to obtain the images of elements of the domain.

Example:

If the functionf squares a real number, thenf(x) =x2_.

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2)

= (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2

=0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0 2 f(−3)

= (−3)2_{−3(−3) + 2 =20}

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0 2 f(−3) = (−3)2−3(−3) + 2

=20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20 3 f(c) +f(1)

= (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20 3 f(c) +f(1) = (c2−3c+ 2)

+ (12−3(1) + 2) =c2−3c+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2)

=c2−3c+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2

−3c+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c

+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2

4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1)

= (c+1)2_{−3(c+1)+2 =c}2_{+2c+1−3c−3+2 =c}2_−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2

=c2_{+2c+1−3c−3+2 =c}2_−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1

−3c−3+2 =c2_−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3

+2 = c2_−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2

=c2_−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c

5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x) h

= [(x+h)

2_{−3(x+h) + 2]−(x}2_{−3x+ 2)}

h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]

−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2)

h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2

−3x−3h+ 2−x2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_−3x−3h

+ 2−x2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2}

−x2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

= h(2x−3 +h) h

Letf(x) = x2_{−3x+ 2.}

1 f(2) = (2)2−3(2) + 2 =0

2 f(−3) = (−3)2−3(−3) + 2 =20

3 f(c) +f(1) = (c2−3c+ 2) + (12−3(1) + 2) =c2−3c+ 2 4 f(c+1) = (c+1)2−3(c+1)+2 =c2+2c+1−3c−3+2 =c2−c 5

f(x+h)−f(x)

h =

[(x+h)2−3(x+h) + 2]−(x2−3x+ 2) h

= x

2_{+ 2xh+h}2_{−3x−3h+ 2−x}2_{+ 3x−2}

h

= 2xh−3h+h

2

h

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1)

= 1 butg is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1

but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Example: Let f(x) = x2,domf =_R and

g(_{x) = x}2_{,domg = [0,+∞). As functions, isf equal to g?}

Solution:

Note thatf(−1) = 1 but g is undefined at −1.

Negative real numbers can’t be inputs tog while f can have any real number as input.

As sets of ordered pairs of real numbers,f contains the ordered pair (−1,1) but g does not.

Thus,f and g are not equal.

Finding the (Natural) Domain of a Function

1 We considerreal-valued functions, meaning functions whose

codomain is _R.

2 If the domain of a function f is not specified, we take it to be

the set of real numbersx such that f(x)∈_R.

1 Denominators should not be zero

Finding the (Natural) Domain of a Function

1 We considerreal-valued functions, meaning functions whose

codomain is _R.

2 If the domain of a function f is not specified, we take it to be

the set of real numbersx such that f(x)∈_R.

1 Denominators should not be zero

Example:

Consider f ={(x, y)∈_R×_R|y = 2x+ 1}. What is domf?

This will be written simply asf(x) = 2x+ 1.

For anyx∈_R, 2x+ 1 is also a real number. Therefore,

Example:

Consider f ={(x, y)∈_R×_R|y = 2x+ 1}. What is domf?

This will be written simply asf(x) = 2x+ 1.

For anyx∈_R, 2x+ 1 is also a real number. Therefore,

Example:

Consider f ={(x, y)∈_R×_R|y = 2x+ 1}. What is domf?

This will be written simply asf(x) = 2x+ 1.

For anyx∈_R,

2x+ 1 is also a real number. Therefore,

Example:

Consider f ={(x, y)∈_R×_R|y = 2x+ 1}. What is domf?

This will be written simply asf(x) = 2x+ 1.

For anyx∈_R, 2x+ 1 is also a real number.

Therefore,

Example:

Consider f ={(x, y)∈_R×_R|y = 2x+ 1}. What is domf?

This will be written simply asf(x) = 2x+ 1.

For anyx∈_R, 2x+ 1 is also a real number. Therefore,

Example: Find the domain of f(x) = 2x x2_−5x.

Solution:

The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0 x= 0,5

Example: Find the domain of f(x) = 2x x2_−5x.

Solution: The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0 x= 0,5

Example: Find the domain of f(x) = 2x x2_−5x.

Solution: The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0

x= 0,5

Example: Find the domain of f(x) = 2x x2_−5x.

Solution: The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0 x= 0,5

Example: Find the domain of f(x) = 2x x2_−5x.

Solution: The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0 x= 0,5

Therefore,domf =_R\ {0,5}.

Example: Find the domain of f(x) = 2x x2_−5x.

Solution: The denominator can’t be zero: x2_{−5x = 0}

x(x−5) = 0 x= 0,5

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0

(x−1)(x−3) ≥ 0 critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞) x−1

− + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞) x−1

− + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞) x−1

− + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞) x−1

− + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞) x−1

− + +

x−3

− − +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 −

+ +

x−3

− − +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − +

+

x−3

− − +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3

− − +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 −

− +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − −

+

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − − +

(x−1)(x−3)

+ − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − − +

(x−1)(x−3) +

− +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − − +

(x−1)(x−3) + −

+

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) =√x2 _{−4x+ 3.}

Solution: The radicand must be nonegative:

x2−4x+ 3 ≥ 0 (x−1)(x−3) ≥ 0

critical points : 1, 3

Table of Signs:

(−∞,1] [1,3) (3,+∞)

x−1 − + +

x−3 − − +

(x−1)(x−3) + − +

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0

2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0

6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x

+ + +

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3

- + + +

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1

- - + +

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+

- +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+

-+

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+ - +

-domf = (−3

Example: Find the domain of f(x) = s − 2 x−1 −

5x (x−1)(2x+ 3)

.

Solution: The radicand must be nonnegative:

− 2

x−1 − 5x

(x−1)(2x+3)

≥0

2

x−1 − 5x

(x−1)(2x+3) ≤0 2(2x+3)−5x

(x−1)(2x+3) ≤0 6−x

(x−1)(2x+3) ≤0

critical numbers :−3 2,1,6

Table of Signs:

(−∞,−3 2) (−

3

2,1) (1,6) (6,+∞)

6−x _{+ + +}

-2x+3 _{- + + +}

x−1 _{- - + +}

+ - +

-domf = (−3

Example: Find the domain of f(x) = √

x+ 6 x2_−5x−4.

Solution:

The radicand must be nonnegative, and the denominator must not be zero.

x+ 6 ≥0 and _R\ {solutions to x2−5x−4 = 0} x ≥ −6 and _R\ {x= 5±

√

25−4(1)(−4)

2(1) }

[−6,∞) ∩ _R\ {5−√41

2 ,

5+√41

2 }

domf = [−6,∞)\ {5−√41 2 ,

Example: Find the domain of f(x) = √

x+ 6 x2_−5x−4.

Solution:

The radicand must be nonnegative, and the denominator must not be zero.

x+ 6 ≥0 and _R\ {solutions to x2−5x−4 = 0}

x ≥ −6 and _R\ {x= 5± √

25−4(1)(−4)

2(1) }

[−6,∞) ∩ _R\ {5−√41

2 ,

5+√41

2 }

domf = [−6,∞)\ {5−√41 2 ,

Example: Find the domain of f(x) = √

x+ 6 x2_−5x−4.

Solution:

The radicand must be nonnegative, and the denominator must not be zero.

x+ 6 ≥0 and _R\ {solutions to x2−5x−4 = 0} x ≥ −6 and _R\ {x= 5±

√

25−4(1)(−4)

2(1) }

[−6,∞) ∩ _R\ {5−√41

2 ,

5+√41

2 }

domf = [−6,∞)\ {5−√41 2 ,

Example: Find the domain of f(x) = √

x+ 6 x2_−5x−4.

Solution:

The radicand must be nonnegative, and the denominator must not be zero.

x+ 6 ≥0 and _R\ {solutions to x2−5x−4 = 0} x ≥ −6 and _R\ {x= 5±

√

25−4(1)(−4)

2(1) }

[−6,∞) ∩ _R\ {5−√41

2 ,

5+√41

2 }

domf = [−6,∞)\ {5−√41 2 ,

Example: Find the domain of f(x) = √

x+ 6 x2_−5x−4.

Solution:

The radicand must be nonnegative, and the denominator must not be zero.

x+ 6 ≥0 and _R\ {solutions to x2−5x−4 = 0} x ≥ −6 and _R\ {x= 5±

√

25−4(1)(−4)

2(1) }

[−6,∞) ∩ _R\ {5−√41

2 ,

5+√41

2 }

domf = [−6,∞)\ {5−√41 2 ,