1.7 - functions
DESCRIPTION
1.7 - Functions. 1.7 - Functions. A function is a relation in which each element of the domain is paired with exactly one element of the range. 1.7 - Functions. A function is a relation in which each element of the domain is paired with exactly one element of the range. 1.7 - Functions. - PowerPoint PPT PresentationTRANSCRIPT
1.7 - Functions
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y
-6
-4 9
-1 -6
1 1
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y
-6
-4 9 Y
-1 -6 E
1 1 S
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y b.
-6
-4 9 Y
-1 -6 E
1 1 S
x y
-3 6
2 5
3 1
2 4
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y b.
-6
-4 9 Y
-1 -6 E
1 1 S
x y
-3 6
2 5
3 1
2 4
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y b.
-6
-4 9 Y
-1 -6 E
1 1 S
x y
-3 6
2 5
3 1
2 4
1.7 - Functions• A function is a relation in which each
element of the domain is paired with exactly one element of the range.
There cannot be an x-value repeated!
Ex.1 Determine if each is a function.
a. X Y b.
-6 NOT A
-4 9 Y FUNC.
-1 -6 E
1 1 S
x y
-3 6
2 5
3 1
2 4
Ex. 2 If f(x) = x2 – 5, find the following:
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5
f(-9)
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5
f(-9)
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) =
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 =
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 = [(4)2 – 5]
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 =[(4)2 – 5]
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 =[(4)2 – 5] + 2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2
= [16 – 5] + 2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2
= [16 – 5] + 2 = 11 + 2
Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9)
f(x) = x2 – 5 f(-9) = (-9)2 – 5
= 81 – 5
f(-9) = 76
b. f(6z)
f(x) = x2 – 5
f(6z) = (6z)2 – 5
= 62·z2 – 5
f(6z) = 36z2 – 5
c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2
= [16 – 5] + 2 = 11 + 2
f(4) + 2 = 13
Ex. 2 If f(x) = x2 – 5, find the following: